Convergence of the d'Alembert series. A necessary and sufficient condition for the convergence of a positive-sign number series. Convergence and divergence of alternating series

Before formulating the sign itself, let's consider an important question:
When should the d'Alembert convergence criterion be used?

The main prerequisites for applying the d'Alembert test are as follows:

1) The common member of the series (“stuffing” of the series) includes some number in the degree, for example, , and so on. Moreover, it does not matter at all where these functions are located, in the numerator or in the denominator - it is important that they are present there.

2) The common term of the series includes the factorial. What is factorial?





…


…

! When using the d'Alembert test, we just have to paint the factorial in detail. As in the previous paragraph, the factorial can be located at the top or bottom of the fraction.

3) If there is a "chain of factors" in the common term of the series, for example, . This case is rare.

Along with powers and (and) factorials, polynomials are often found in the filling of the series, this does not change things - you need to use the d'Alembert test.

In addition, in the general term of the series, both the degree and the factorial can occur at the same time; there may be two factorials, two degrees, it is important that there be at least something of the points considered - and this is just a prerequisite for using the d'Alembert sign.

Sign of d'Alembert: Consider positive number series. If there is a limit of the ratio of the next term to the previous one: , then:
a) At a row converges
b) At a row diverges
c) When sign does not respond. You need to use another sign. Most often, a unit is obtained in the case when they try to apply the d'Alembert test where it is necessary to use the limit comparison test.

Without an understanding of the limit and the ability to reveal the uncertainty further, unfortunately, one cannot move forward.

Example:
Decision: We see that in the common term of the series we have , and this is the correct premise that we need to use the d'Alembert test.

We use the d'Alembert sign:


converges.

Cauchy's radical sign.

The Cauchy convergence test for positive numerical series is somewhat similar to the d'Alembert test just considered.

Cauchy's radical sign: Consider positive number series. If there is a limit: , then:
a) At a row converges. In particular, the series converges for .
b) At a row diverges. In particular, the series diverges at .
c) When sign does not respond. You need to use another sign.

! It is interesting to note that if the Cauchy test does not give us an answer to the question of the convergence of the series, then the d'Alembert test will not give us an answer either. But if d'Alembert's sign does not give an answer, then Cauchy's sign may well "work". That is, the Cauchy sign is in this sense a stronger sign.

!!! When should you use the Cauchy radical sign? The radical Cauchy test is usually used in cases where the common term of the series FULLY is in the degree dependent on "en". Or when the root "good" is extracted from the common member of the series. There are still exotic cases, but we will not hammer our heads with them.

Example: Examine the series for convergence

Decision: We see that the common term of the series is completely under the degree depending on , which means that we need to use the radical Cauchy criterion:


Thus, the series under study diverges.

Integral Cauchy test.

In order to apply the Cauchy integral criterion, it is necessary to more or less confidently be able to find derivatives, integrals, and also have the skill of calculating improper integral first kind.

I will formulate in my own words (for ease of understanding).

Integral Cauchy sign: Consider positive number series. This series converges or diverges together with the corresponding improper integral.

! !! The main prerequisite for using the integral Cauchy test is the fact that the common member of the series contains some function and its derivative.

Example: Examine the series for convergence

Decision: From the topic Derivative you probably remember the simplest tabular thing: , and we have just such a canonical case.

How to use the integral sign? First, we take the integral icon and rewrite the upper and lower limits from the “counter” of the row: . Then, under the integral, we rewrite the “stuffing” of the series with the letter “x”:.

Now we need to calculate the improper integral. In this case, two cases are possible:

1) If it turns out that the integral converges, then our series will also converge.

2) If it turns out that the integral diverges, then our series will also diverge.

We use the integral feature:

The integrand is continuous on

Thus, the series under study diverges together with the corresponding improper integral.

Example: Investigate the convergence of a series

Decision: First of all, we check a necessary criterion for the convergence of the series. This is not a formality, but a great chance to deal with the example of "little bloodshed".

Numeric sequence higher order of growth than , therefore , that is, the necessary criterion for convergence is satisfied, and the series can both converge and diverge.

Thus, some sign must be used. But what? Limit sign of comparison clearly does not fit, since the logarithm has been tucked into the common term of the series, signs of d'Alembert and Cauchy also do not lead to results. If we had, then at the very least it would be possible to wriggle out through integral feature.

"Inspection of the scene" suggests a divergent series (the case of a generalized harmonic series), but again the question arises, how to take into account the logarithm in the numerator?

There remains the very first sign of comparison, based on inequalities, which is often not taken into account and gathers dust on the far shelf. Let's write a series in more detail:

I remind you that - unlimited growing numerical sequence:

And, starting from the number , the inequality will be fulfilled:

that is, the members of the series will be much more relevant members divergent row .

As a result, there is nothing left for the series but to diverge as well.

The convergence or divergence of a numerical series depends on its "infinite tail" (remainder). In our case, we can ignore the fact that the inequality is not true for the first two numbers - this does not affect the conclusion.

The clean design of the example should look something like this:

Compare this series with the divergent series .
For all numbers, starting from , the inequality is satisfied, therefore, by comparison, the series under study diverges.

Alternating rows. Leibniz sign. Solution examples.

What is an alternating series? This is clear or almost clear already from the name itself. Just the simplest example.

Consider the series and write it in more detail:

Alternation provides a multiplier: if even, then there will be a plus sign, if odd, a minus sign

In practical examples, the alternation of the terms of the series can provide not only the factor , but also its brothers: , , , …. For example:

The pitfall is "tricks":,, etc. are such multipliers do not provide sign change. It is quite clear that for any natural : , , .

How to examine an alternating series for convergence? Use the Leibniz sign.

Leibniz sign: If two conditions are met in the alternating series: 1) the terms of the series monotonically decrease in absolute value. 2) the limit of the common term is equal to zero in absolute value, then the series converges, and the modulus of the sum of this series does not exceed the modulus of the first term.

Brief information about the module:

What does "modulo" mean? The module, as we remember from school, "eats" the minus sign. Let's go back to the series . Mentally erase all signs with an eraser and look at the numbers. We will see that each next row member smaller than the previous one.

Now a little about monotony.

Row members strictly monotone decrease modulo if EACH NEXT member of the series modulo LESS than previous: . For a number strict monotonicity of decreasing is fulfilled, it can be described in detail:

And we can say in short: each next member of the series modulo less than the previous one: .

Row members not strictly monotone decrease in modulus, if EACH NEXT term of the series modulo IS NOT GREATER THAN the previous one: . Consider a series with a factorial: Here, non-strict monotonicity takes place, since the first two terms of the series are identical in absolute value. That is, each next member of the series modulo no more than the previous one: .

Under the conditions of Leibniz's theorem, the monotonicity of the decrease must be satisfied (it doesn't matter if it is strict or non-strict). In this case, the members of the series can even increase modulo for some time, but the "tail" of the series must necessarily be monotonically decreasing.

Example: Examine the series for convergence

Decision: The common term of the series includes the factor , which means that you need to use the Leibniz test

1) Checking the series for monotonic decrease.

1<2<3<…, т.е. n+1>n the first condition is not met

2) – the second condition is also not met.

Conclusion: the series diverges.

Definition: If a series converges according to the Leibniz criterion and a series composed of modules also converges, then we say that the series converges absolutely.

If the series converges according to the Leibniz criterion, and the series composed of modules diverges, then the series is said to be converges conditionally.

If a series composed of modules converges, then this series also converges.

Therefore, an alternating convergent series must be examined for absolute or conditional convergence.

Example:

Decision: We use the Leibniz sign:

1) Each next member of the series is less in modulus than the previous one: – the first condition is met.

2) – the second condition is also fulfilled.

Conclusion: the series converges.

Check for conditional or absolute convergence.

Let's make a series of modules - again we just remove the multiplier, which ensures the alternation:
- diverges (harmonic series).

Thus, our series is not absolutely convergent.
Study Series converges conditionally.

Example: Examine a series for conditional or absolute convergence

Decision: We use the Leibniz sign:
1) Let's try to write down the first few terms of the series:


…?!

2)

The fact is that there are no standard everyday tricks for solving such limits. Where does this limit go? To zero, to infinity? It is important here that WHAT grows faster at infinity- numerator or denominator.

If the numerator at grows faster than the factorial, then . If, at infinity, the factorial grows faster than the numerator, then, on the contrary, it “pulls” the limit to zero: . Or maybe this limit is equal to some non-zero number? or . Instead, you can substitute some polynomial of a thousandth degree, this again will not change the situation - sooner or later the factorial will still “overtake” such a terrible polynomial. Factorial higher order of growth.

– The factorial grows faster than product of any quantity exponential and power sequences(our case).

– Any exponential sequence grows faster than any power sequence, for example: , . exponential sequence higher order of growth than any power sequence. Similar to the factorial, the exponential sequence "pulls" the product of any number of any power sequences or polynomials: .

– Is there anything “stronger” than the factorial? There is! The exponential sequence ("en" to the power of "en") grows faster than the factorial. In practice, it is rare, but the information will not be superfluous.

End of help

Thus, the second point of the study can be written as follows:
2) , since a higher order of growth than .
The terms of the series decrease modulo, starting from some number, at the same time, each next term of the series is less in absolute value than the previous one, thus, the decrease is monotonous.

Conclusion: the series converges.

Here is just the curious case when the terms of the series first grow in absolute value, which is why we have an erroneous initial opinion about the limit. But, starting from some number "en", the factorial overtakes the numerator, and the “tail” of the series becomes monotonically decreasing, which is fundamentally important for fulfilling the conditions of the Leibniz theorem. What exactly is this "en" is quite difficult to find out.

We examine the series for absolute or conditional convergence:

And here the d'Alembert sign is already working:

We use the d'Alembert sign:

Thus, the series converges.

Study Series converges absolutely.

The analyzed example can be solved in another way (we use a sufficient criterion for the convergence of an alternating series).

A sufficient criterion for the convergence of an alternating series is: If a series composed of the absolute values ​​of the members of a given series converges, then the given series also converges.

Second way:

Examine a series for conditional or absolute convergence

Decision : We examine the series for absolute convergence:

We use the d'Alembert sign:

Thus, the series converges.
Based on a sufficient criterion for the convergence of an alternating series, the series itself converges.

Conclusion: Study series converges absolutely.

To calculate the sum of a series with a given accuracy we will use the following theorem:

Let the alternating series satisfies the conditions of the Leibniz test and let - its n-th partial sum. Then the series converges and the error in the approximate calculation of its sum S in absolute value does not exceed the modulus of the first discarded term:

functional rows. Power series.
The region of convergence of the series.

To successfully master the topic, you need to be well versed in ordinary numerical series.

If a positive series has a limit of the form
, then this series converges at
and diverges at
. At
d'Alembert's test does not solve the question of the convergence of the series.

Comment. The d'Alembert test is used if the formula for the common term of the series contains a factor or is an exponential function of n.

Example 1
.

Decision. The possibility of using the d'Alembert sign is indicated by the presence in the formula of a common term of factors and .

Here
. To receive
replace in the formula all n on the n+1. Get
. Then

According to the d'Alembert test, the series under study converges.

Example 2 Investigate for convergence series
.

Decision. Here
,
. We apply the d'Alembert sign:

The series diverges.

Investigate for convergence, using the d'Alembert test, the following series:

a)
; b)
; in)
; G)
.

2.3.4. Cauchy's radical sign

If for a positive series there is a limit
, then at l<1 данный ряд сходится, а при l> 1 − diverges. At l=1 the radical Cauchy test, as well as the d'Alembert test, does not answer the question about the convergence of the series.

In practice, the Cauchy criterion is most often used when the common term of the series is an exponential or exponential-power function of n.

Example 1 Investigate for convergence series

.

Decision. The common term of the series contains an expression in the power n. Therefore, it is advisable to use the radical Cauchy test:

.

The series under study converges.

Example 2 Investigate for convergence series
.

Decision. Calculate the limit

where
.

To calculate this limit, we use the L'Hospital rule, having previously taken the logarithm of the resulting expression:

.

Since ln t=0, then t=1, and, therefore,
.

As
, then the series diverges by the radical Cauchy criterion.

Examples for self-solving

Investigate for convergence using the radical Cauchy test the following series:

a)
; b)
; in)
; G)
.

2.3.5. Integral Cauchy test

Let the common term of the series be the value of the function
at
, i.e.
. If at the same time the function
monotonically decreases in some interval
, where
, then this series converges if the improper integral converges
, and diverges if this improper integral diverges. An important practical consequence follows from this theorem: for a convergent series with a common term that satisfies the conditions of the theorem, the remainder of the series can be estimated from the relation
.

Consider examples of the application of the Cauchy integral test.

Example. Investigate for convergence series
.

Decision. As
is the function value
at
and this function is continuous and decreases monotonically in the interval
, then we examine the convergence of the improper integral

The integral diverges. Therefore, this series also diverges.

Examples for independent decision.

Investigate for convergence using the integral Cauchy test the following series:

a)
; b)
; in)
; G)
.

2.4. Convergence and divergence of alternating series

Literature:, Part 3, Ch. 15, § 15.4

Row with members of arbitrary signs is called alternating. In what follows, series with an infinite number of positive and negative terms will be considered.

alternating series called absolutely convergent, if the series composed of the moduli of its terms converges
.

Theorem on the absolute convergence of a series: if a series composed of modules of the terms of this series converges
, then this alternating series also converges (i.e., is absolutely convergent).

Row called conditionally convergent, if it converges, and the series composed of the moduli of its members
, diverges.

Basic properties of absolutely and conditionally convergent series:

1) an absolutely convergent series remains convergent and does not change the value of the sum for any permutation of its members;

2) by changing the order of the terms in a conditionally convergent series, one can make the sum of the series equal to any predetermined number or even make the series divergent;

3) if an alternating series converges absolutely, then series composed of only its positive or only negative terms converge; if the series converges conditionally, then the series mentioned above converge.

Example 1

Decision. The series is alternating, in connection with the growth of the denominator of its members, the latter decrease in absolute value. It is easy to see that the limit of the common term of the series at n→∞ is zero:
. Hence, according to the Leibniz criterion, this alternating series converges.

Consider now a series composed of modules of its members,

This is a harmonic series that is known to diverge. Therefore, this series is conditionally convergent.

Example 2 Examine the series for absolute and conditional convergence

Decision. This series is sign-changing, since
can take both positive and negative values. Consider a series composed of modules of members of this series:

In view of the obvious inequality
by the comparison test for positive series, we have that the series
converges, since series converges
. From the convergence of the series
by the criterion of absolute convergence, we have that the series
converges and, moreover, absolutely.

This article has collected and structured the information necessary to solve almost any example on the topic of number series, from finding the sum of a series to examining its convergence.

Article review.

Let's start with the definitions of a positive-sign, alternating-sign series and the concept of convergence. Next, consider standard series, such as a harmonic series, a generalized harmonic series, and recall the formula for finding the sum of an infinitely decreasing geometric progression. After that, we turn to the properties of convergent series, dwell on the necessary condition for the convergence of the series, and state sufficient criteria for the convergence of the series. We will dilute the theory by solving typical examples with detailed explanations.

Page navigation.

Basic definitions and concepts.

Let we have a numerical sequence , where .

Here is an example of a numerical sequence: .

Number series is the sum of members of a numerical sequence of the form .

As an example of a number series, we can give the sum of an infinitely decreasing geometric progression with the denominator q = -0.5: .

are called common member of the number series or the kth member of the series.

For the previous example, the common term of the number series is .

Partial sum of a number series is a sum of the form , where n is some natural number. also called the n-th partial sum of the number series.

For example, the fourth partial sum of the series there is .

Partial sums form an infinite sequence of partial sums of a numerical series.

For our series, the nth partial sum is found by the formula for the sum of the first n terms of a geometric progression , that is, we will have the following sequence of partial sums: .

The number line is called converging, if there is a finite limit of the sequence of partial sums . If the limit of the sequence of partial sums of a numerical series does not exist or is infinite, then the series is called divergent.

The sum of a convergent number series is called the limit of the sequence of its partial sums, that is, .

In our example, therefore, the series converges, and its sum is equal to sixteen thirds: .

An example of a divergent series is the sum of a geometric progression with a denominator greater than one: . The nth partial sum is given by , and the limit of partial sums is infinite: .

Another example of a divergent number series is the sum of the form . In this case, the nth partial sum can be calculated as . The limit of partial sums is infinite .

Sum view called harmonic number series.

Sum view , where s is some real number, is called generalized harmonic number series.

The above definitions are sufficient to substantiate the following very frequently used statements, we recommend that you remember them.

THE HARMONIC SERIES IS Divergent.

Let us prove the divergence of the harmonic series.

Let's assume that the series converges. Then there is a finite limit of its partial sums. In this case, we can write and , which leads us to the equality .

On the other side,


The following inequalities are beyond doubt. Thus, . The resulting inequality tells us that the equality cannot be achieved, which contradicts our assumption about the convergence of the harmonic series.

Conclusion: the harmonic series diverges.

THE SUMMATION OF A GEOMETRIC PROGRESSION OF THE TYPE WITH A DENOMINATOR q IS A CONVERGENT NUMERICAL SERIES IF , AND A DIVERGENT SERIES AT .

Let's prove it.

We know that the sum of the first n terms of a geometric progression is found by the formula .

When fair


which indicates the convergence of the numerical series.

For q = 1 we have a number series . Its partial sums are found as , and the limit of partial sums is infinite , which indicates the divergence of the series in this case.

If q \u003d -1, then the number series will take the form . Partial sums take on a value for odd n , and for even n . From this we can conclude that the limit of partial sums does not exist and the series diverges.

When fair


which indicates the divergence of the numerical series.

GENERALIZED HARMONIC SERIES CONVERGES FOR s > 1 AND DIVERS FOR .

Proof.

For s = 1 we get the harmonic series , and above we have established its divergence.

At s the inequality holds for all natural k . Due to the divergence of the harmonic series, it can be argued that the sequence of its partial sums is unlimited (since there is no finite limit). Then the sequence of partial sums of the number series is all the more unlimited (each member of this series is greater than the corresponding member of the harmonic series), therefore, the generalized harmonic series diverges at s.

It remains to prove the convergence of the series for s > 1 .

Let's write the difference:



Obviously, then


Let's write the resulting inequality for n = 2, 4, 8, 16, …



Using these results, the following actions can be performed with the original numerical series:



Expression is the sum of a geometric progression whose denominator is . Since we are considering the case for s > 1, then . So
. Thus, the sequence of partial sums of the generalized harmonic series for s > 1 is increasing and at the same time bounded from above by the value , therefore, it has a limit, which indicates the convergence of the series . The proof is complete.

The number line is called sign-positive if all its terms are positive, that is, .

The number line is called alternating if the signs of its neighboring terms are different. An alternating number series can be written as or , where .

The number line is called alternating if it contains an infinite number of both positive and negative terms.

An alternating number series is a special case of an alternating series.

ranks

are sign-positive, sign-alternating, and sign-alternating, respectively.

For an alternating series, there is the concept of absolute and conditional convergence.

absolutely convergent, if a series of absolute values ​​of its members converges, that is, a positive-sign numerical series converges.

For example, number lines and absolutely converge, since the series converges , which is the sum of an infinitely decreasing geometric progression.

The alternating series is called conditionally convergent if the series diverges and the series converges.

An example of a conditionally convergent number series is the series . Number series , composed of the absolute values ​​of the members of the original series, divergent, since it is harmonic. At the same time, the original series is convergent, which is easily established using . Thus, the numerical sign-alternating series conditionally convergent.

Properties of convergent numerical series.

Example.

Prove the convergence of the numerical series.

Decision.

Let's write the series in a different form . The number series converges, since the generalized harmonic series is convergent for s > 1, and due to the second property of convergent number series, the series with the numerical coefficient will also converge.

Example.

Does the number series converge?

Decision.

Let's transform the original series: . Thus, we have obtained the sum of two numerical series and , and each of them converges (see the previous example). Therefore, due to the third property of convergent numerical series, the original series also converges.

Example.

Prove the convergence of the number series and calculate its sum.

Decision.

This number series can be represented as the difference of two series:

Each of these series is the sum of an infinitely decreasing geometric progression, therefore, is convergent. The third property of convergent series allows us to assert that the original numerical series converges. Let's calculate its sum.

The first term of the series is one, and the denominator of the corresponding geometric progression is 0.5, therefore, .

The first term of the series is 3, and the denominator of the corresponding infinitely decreasing geometric progression is 1/3, so .

Let's use the obtained results to find the sum of the original number series:

A necessary condition for the convergence of a series.

If the number series converges, then the limit of its k-th term is equal to zero: .

When examining any numerical series for convergence, one should first of all check the fulfillment of the necessary convergence condition. Failure to comply with this condition indicates the divergence of the numerical series, that is, if , then the series diverges.

On the other hand, it must be understood that this condition is not sufficient. That is, the fulfillment of equality does not indicate the convergence of the numerical series. For example, for a harmonic series, the necessary convergence condition is satisfied, and the series diverges.

Example.

Examine the number series for convergence.

Decision.

Let's check the necessary condition for the convergence of the numerical series:

Limit n-th member of the numerical series is not equal to zero, therefore, the series diverges.

Sufficient conditions for the convergence of a positive sign series.

When using sufficient features to study numerical series for convergence, you constantly have to deal with , so we recommend that you refer to this section in case of difficulty.

A necessary and sufficient condition for the convergence of a positive-sign number series.

For the convergence of a sign-positive number series it is necessary and sufficient that the sequence of its partial sums be bounded.

Let's start with series comparison features. Their essence lies in comparing the studied numerical series with a series whose convergence or divergence is known.

First, second and third signs of comparison.

The first sign of comparison of rows.

Let and be two positive-sign numerical series and the inequality holds for all k = 1, 2, 3, ... Then the convergence of the series implies the convergence , and the divergence of the series implies the divergence .

The first comparison criterion is used very often and is a very powerful tool for examining numerical series for convergence. The main problem is the selection of a suitable series for comparison. The series for comparison is usually (but not always) chosen so that the exponent of its k-th member is equal to the difference between the exponents of the numerator and denominator of the k-th member of the number series under study. For example, let, the difference between the exponents of the numerator and denominator is 2 - 3 = -1, therefore, for comparison, we select a series with the kth member, that is, a harmonic series. Let's look at a few examples.

Example.

Set the convergence or divergence of the series.

Decision.

Since the limit of the common term of the series is equal to zero, then the necessary condition for the convergence of the series is satisfied.

It is easy to see that the inequality is true for all natural k . We know that the harmonic series diverges, therefore, according to the first sign of comparison, the original series is also divergent.

Example.

Examine the number series for convergence.

Decision.

The necessary condition for the convergence of the number series is satisfied, since . It is obvious that the inequality for any natural value of k. The series converges because the generalized harmonic series converges for s > 1. Thus, the first sign of series comparison allows us to state the convergence of the original numerical series.

Example.

Determine the convergence or divergence of the number series.

Decision.

, therefore, the necessary condition for the convergence of the numerical series is satisfied. Which row to choose for comparison? A numerical series suggests itself, and in order to determine s, we carefully examine the numerical sequence. The terms of the numerical sequence increase towards infinity. Thus, starting from some number N (namely, from N = 1619 ), the terms of this sequence will be greater than 2 . Starting from this number N , the inequality is valid . The number series converges due to the first property of convergent series, since it is obtained from a convergent series by discarding the first N - 1 terms. Thus, according to the first sign of comparison, the series is convergent, and due to the first property of convergent numerical series, the series will also converge.

The second sign of comparison.

Let and be sign-positive numerical series. If , then the convergence of the series implies the convergence of . If , then the divergence of the numerical series implies the divergence of .

Consequence.

If and , then the convergence of one series implies the convergence of the other, and the divergence implies divergence.

We examine the series for convergence using the second comparison criterion. Let's take a convergent series as a series. Let's find the limit of the ratio of the k-th members of the numerical series:

Thus, according to the second criterion of comparison, the convergence of the numerical series implies the convergence of the original series.

Example.

Investigate the convergence of a number series.

Decision.

Let us check the necessary condition for the convergence of the series . The condition is met. To apply the second sign of comparison, let's take a harmonic series. Let's find the limit of the ratio of k-th terms:

Consequently, the divergence of the original series follows from the divergence of the harmonic series according to the second criterion of comparison.

For information, we present the third criterion for comparing series.

The third sign of comparison.

Let and be sign-positive numerical series. If the condition is satisfied from a certain number N, then the convergence of the series implies the convergence, and the divergence of the series implies the divergence.

Sign of d'Alembert.

Comment.

d'Alembert's sign is valid if the limit is infinite, that is, if , then the series converges if , then the series diverges.

If , then the d'Alembert test does not provide information about the convergence or divergence of the series, and additional research is required.

Example.

Examine the number series for convergence on the basis of d'Alembert.

Decision.

Let's check the fulfillment of the necessary condition for the convergence of the numerical series, we calculate the limit by:

The condition is met.

Let's use d'Alembert's sign:

Thus, the series converges.

Cauchy's radical sign.

Let be a positive sign number series. If , then the series converges, if , then the series diverges.

Comment.

Cauchy's radical test is valid if the limit is infinite, that is, if , then the series converges if , then the series diverges.

If , then the radical Cauchy test does not provide information about the convergence or divergence of the series and additional research is required.

It is usually easy enough to see the cases where it is best to use the radical Cauchy test. A characteristic case is when the common term of the numerical series is an exponential power expression. Let's look at a few examples.

Example.

Investigate a positive-sign number series for convergence using the radical Cauchy test.

Decision.

. By the radical Cauchy test, we get .

Therefore, the series converges.

Example.

Does the number series converge? .

Decision.

Let's use the radical Cauchy test , therefore, the number series converges.

Integral Cauchy test.

Let be a positive sign number series. Let us compose a function of continuous argument y = f(x) , similar to the function . Let the function y = f(x) be positive, continuous and decreasing on the interval , where ). Then in case of convergence improper integral converges the studied number series. If the improper integral diverges, then the original series also diverges.

When checking the decay of a function y = f(x) over an interval, you may find the theory in the section useful.

Example.

Examine the number series with positive terms for convergence.

Decision.

The necessary condition for the convergence of the series is satisfied, since . Let's consider a function. It is positive, continuous and decreasing on the interval . The continuity and positivity of this function is beyond doubt, but let us dwell on the decrease in a little more detail. Let's find the derivative:
. It is negative on the interval , therefore, the function decreases on this interval.

Before starting work with this topic, I advise you to look at the section with terminology for number series. It is especially worth paying attention to the concept of a common term of a series. If you have doubts about the correct choice of the sign of convergence, I advise you to look at the topic "Choosing the sign of convergence of numerical series".

The D'Alembert test (or d'Alembert test) is used to study the convergence of series whose common term is strictly greater than zero, i.e. $u_n > 0$. Such series are called strictly positive. In standard examples, the sign of D "Alembert is used in the limiting form.

Sign of D "Alamber (in the limiting form)

If the series $\sum\limits_(n=1)^(\infty)u_n$ is strictly positive and $$ \lim_(n\to\infty)\frac(u_(n+1))(u_n)=L, $ $ then for $L<1$ ряд сходится, а при $L>1$ (and for $L=\infty$) the series diverges.

The formulation is quite simple, but the following question remains open: what happens if $L=1$? The sign of D "Alembert is not able to answer this question. If $L \u003d 1 $, then the series can both converge and diverge.

Most often, in standard examples, the sign of D "Alembert is used if the expression of the common term of the series contains a polynomial in $n$ (the polynomial can also be under the root) and a degree of the form $a^n$ or $n!$. For example, $u_n= \frac(5^n\cdot(3n+7))(2n^3-1)$ (see example #1) or $u_n=\frac(\sqrt(4n+5))((3n-2)$ (см. пример №2). Вообще, для стандартного примера наличие $n!$ - это своеобразная "визитная карточка" признака Д"Аламбера.!}

What does the expression "n!" stand for? show/hide

Recording "n!" (read "en factorial") denotes the product of all natural numbers from 1 to n, i.e.

$$ n!=1\cdot2\cdot 3\cdot \ldots\cdot n $$

By definition, it is assumed that $0!=1!=1$. For example, let's find 5!:

$$ 5!=1\cdot 2\cdot 3\cdot 4\cdot 5=120. $$

In addition, the D "Alembert test is often used to determine the convergence of a series whose common term contains the product of the following structure: $u_n=\frac(3\cdot 5\cdot 7\cdot\ldots\cdot(2n+1))(2\ cdot 5\cdot 8\cdot\ldots\cdot(3n-1))$.

Example #1

Examine the series $\sum\limits_(n=1)^(\infty)\frac(5^n\cdot(3n+7))(2n^3-1)$ for convergence.

Since the lower summation limit is equal to 1, the common term of the series is written under the sum sign: $u_n=\frac(5^n\cdot(3n+7))(2n^3-1)$. Since for $n≥ 1$ we have $3n+7 > 0$, $5^n>0$ and $2n^3-1 > 0$, then $u_n > 0$. Therefore, our series is strictly positive.

$$ 5\cdot\lim_(n\to\infty)\frac((3n+10)\left(2n^3-1\right))(\left(2(n+1)^3-1\right )(3n+7))=\left|\frac(\infty)(\infty)\right|= 5\cdot\lim_(n\to\infty)\frac(\frac((3n+10)\left (2n^3-1\right))(n^4))(\frac(\left(2(n+1)^3-1\right)(3n+7))(n^4))= 5 \cdot\lim_(n\to\infty)\frac(\frac(3n+10)(n)\cdot\frac(2n^3-1)(n^3))(\frac(\left(2( n+1)^3-1\right))(n^3)\cdot\frac(3n+7)(n))=\\ =5\cdot\lim_(n\to\infty)\frac(\ left(\frac(3n)(n)+\frac(10)(n)\right)\cdot\left(\frac(2n^3)(n^3)-\frac(1)(n^3) \right))(\left(2\left(\frac(n)(n)+\frac(1)(n)\right)^3-\frac(1)(n^3)\right)\cdot \left(\frac(3n)(n)+\frac(7)(n)\right))=5\cdot\lim_(n\to\infty)\frac(\left(3+\frac(10) (n)\right)\cdot\left(2-\frac(1)(n^3)\right))(\left(2\left(1+\frac(1)(n)\right)^3 -\frac(1)(n^3)\right)\cdot\left(3+\frac(7)(n)\right))=5\cdot\frac(3\cdot 2)(2\cdot 3 )=5. $$

Since $\lim_(n\to\infty)\frac(u_(n+1))(u_n)=5>1$, then according to the given series diverges.

To be honest, the sign of D "Alembert is not the only option in this situation. You can use, for example, the radical Cauchy sign. However, the use of the radical Cauchy sign will require knowledge (or proof) of additional formulas. Therefore, using the sign of D" Alembert in this situation is more convenient.

Answer: the series diverges.

Example #2

Explore the series $\sum\limits_(n=1)^(\infty)\frac(\sqrt(4n+5))((3n-2)$ на сходимость.!}

Since the lower summation limit is 1, the common term of the series is written under the sum sign: $u_n=\frac(\sqrt(4n+5))((3n-2)$. Заданный ряд является строго положительным, т.е. $u_n>0$.!}

The common term of the series contains a polynomial under the root, i.e. $\sqrt(4n+5)$, and factorial $(3n-2)!$. The presence of a factorial in a standard example is an almost one hundred percent guarantee of the application of the D "Alembert sign.

To apply this feature, we have to find the limit of the relation $\frac(u_(n+1))(u_n)$. To write $u_(n+1)$, you need to use the formula $u_n=\frac(\sqrt(4n+5))((3n-2)$ вместо $n$ подставить $n+1$:!}

$$ u_(n+1)=\frac(\sqrt(4(n+1)+5))((3(n+1)-2)=\frac{\sqrt{4n+9}}{(3n+1)!}. $$ !}

Since $(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$, the formula for $u_(n+1)$ can be written as otherwise:

$$ u_(n+1)=\frac(\sqrt(4n+9))((3n+1)=\frac{\sqrt{4n+9}}{(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)}. $$ !}

This entry is convenient for further solution when we have to reduce the fraction under the limit. If equality with factorials requires clarification, then please expand the note below.

How did we get $(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$? show/hide

The notation $(3n+1)!$ means the product of all natural numbers from 1 to $3n+1$. Those. this expression can be written like this:

$$ (3n+1)!=1\cdot 2\cdot\ldots\cdot(3n+1). $$

Immediately before the number $3n+1$ there is a number one less, i.e. number $3n+1-1=3n$. And immediately before the number $3n$ is the number $3n-1$. Well, just before the number $3n-1$ we have the number $3n-1-1=3n-2$. Let's rewrite the formula for $(3n+1)!$:

$$ (3n+1)!=1\cdot2\cdot\ldots\cdot(3n-2)\cdot(3n-1)\cdot 3n\cdot (3n+1) $$

What is the product of $1\cdot2\cdot\ldots\cdot(3n-2)$? This product is equal to $(3n-2)!$. Therefore, the expression for $(3n+1)!$ can be rewritten in this form:

$$(3n+1)!=(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)$$

This entry is convenient for further solution when we have to reduce the fraction under the limit.

Calculate the value of $\lim_(n\to\infty)\frac(u_(n+1))(u_n)$:

$$ \lim_(n\to\infty)\frac(u_(n+1))(u_n)=\lim_(n\to\infty)\frac(\frac(\sqrt(4n+9))(( 3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)))(\frac(\sqrt(4n+5))((3n-2)}= \lim_{n\to\infty}\left(\frac{\sqrt{4n+9}}{\sqrt{4n+5}}\cdot\frac{(3n-2)!}{(3n-2)!\cdot (3n-1)\cdot 3n\cdot(3n+1)}\right)=\\ =\lim_{n\to\infty}\frac{\sqrt{4n+9}}{\sqrt{4n+5}}\cdot\lim_{n\to\infty}\frac{1}{(3n-1)\cdot 3n\cdot(3n+1)}= \lim_{n\to\infty}\frac{\sqrt{4+\frac{9}{n}}}{\sqrt{4+\frac{5}{n}}}\cdot\lim_{n\to\infty}\frac{1}{(3n-1)\cdot 3n\cdot(3n+1)}=1\cdot 0=0. $$ !}

Since $\lim_(n\to\infty)\frac(u_(n+1))(u_n)=0<1$, то согласно

d'Alembert convergence test

Jean Léron d'Alembert is a famous French mathematician of the 18th century. In general, d'Alembert specialized in differential equations and, on the basis of his research, was engaged in ballistics, so that His Majesty's cannonballs would fly better. At the same time, I didn’t forget about the numerical series, it was not for nothing that the ranks of the Napoleonic troops converged and diverged so clearly.

Before formulating the sign itself, let's consider an important question: When should the d'Alembert convergence criterion be used?

Let's start with repetition first. Recall the cases when you need to use the most popular marginal comparison criterion. The limit sign of comparison is used when in the common term of the series: 1) There is a polynomial in the denominator. 2) Polynomials are in both the numerator and the denominator. 3) One or both polynomials can be under the root.

The main prerequisites for applying the d'Alembert sign are as follows:

1) The common member of the series ("stuffing" of the series) includes some number in the degree, for example, and so on. Moreover, it does not matter at all where this thing is located, in the numerator or in the denominator - it is important that it is present there.

2) The common term of the series includes the factorial. What is factorial? Nothing complicated, the factorial is just a folded record of the product: ......

! When using the d'Alembert test, we just have to paint the factorial in detail. As in the previous paragraph, the factorial can be located at the top or bottom of the fraction.

3) If there is a “chain of factors” in the common term of the series, for example, . This case is rare, but! When studying such a series, a mistake is often made - see Example 6.

Along with powers and (and) factorials, polynomials are often found in the filling of the series, this does not change things - you need to use the d'Alembert test.

In addition, in the general term of the series, both the degree and the factorial can occur at the same time; there may be two factorials, two degrees, it is important that there be at least something of the points considered - and this is just a prerequisite for using the d'Alembert sign.

Sign of d'Alembert: Consider positive number series. If there is a limit to the ratio of the next term to the previous:, then: a) Order convergesdivergessign does not respond. You need to use another sign. Most often, a unit is obtained in the case when they try to apply the d'Alembert test where it is necessary to use the limit comparison test.

If you still have problems with limits or misunderstanding of limits, please refer to the lesson Limits. Solution examples. Without an understanding of the limit and the ability to reveal the uncertainty further, unfortunately, one cannot move forward.

Cauchy's radical sign

Augustin Louis Cauchy is an even more famous French mathematician. Any student of a technical specialty can tell you Cauchy's biography. In the most beautiful colors. It is no coincidence that this surname is carved on the first floor of the Eiffel Tower.

The Cauchy convergence test for positive numerical series is somewhat similar to the d'Alembert test just considered.

Cauchy's radical sign: Consider positive number series. If there is a limit: then: a) Order converges. In particular, the series converges for . b) Order diverges. In particular, the series diverges at . c) When sign does not respond. You need to use another sign. It is interesting to note that if the Cauchy test does not give us an answer to the question of the convergence of the series, then the d'Alembert test will not give us an answer either. But if d'Alembert's sign does not give an answer, then Cauchy's sign may well "work". That is, the Cauchy sign is in this sense a stronger sign.

When should you use the Cauchy radical sign? The radical Cauchy test is usually used in cases where the common term of the series FULLY is in the degree dependent on "en". Or when the root "good" is extracted from the common member of the series. There are still exotic cases, but we will not hammer our heads with them.