Example No. 9

Investigate the convergence of the series $\sum\limits_(n=1)^(\infty)\frac(1)(\sqrt(n))\arctg\frac(\pi)(\sqrt(2n-1))$.

Since the lower limit of summation is 1, the general term of the series is written under the sum sign: $u_n=\frac(1)(\sqrt(n))\arctg\frac(\pi)(\sqrt(2n-1))$ . First, let's determine whether this series is positive, i.e. Is the inequality $u_n≥ 0$ true? The factor $\frac(1)(\sqrt(n))> 0$, this is clear, but what about the arctangent? There is nothing complicated with the arctange: since $\frac(\pi)(\sqrt(2n-1)) >0$, then $\arctg\frac(\pi)(\sqrt(2n-1))>0$ . Conclusion: our series is positive. Let us apply the comparison criterion to study the issue of convergence of this series.

First, let's select a series with which we will compare. If $n\to\infty$, then $\frac(\pi)(\sqrt(2n-1))\to 0$. Therefore, $\arctg\frac(\pi)(\sqrt(2n-1))\sim\frac(\pi)(\sqrt(2n-1))$. Why is that? If we look at the table at the end of this document, we will see the formula $\arctg x\sim x$ for $x\to 0$. We used this formula, only in our case $x=\frac(\pi)(\sqrt(2n-1))$.

In the expression $\frac(1)(\sqrt(n))\arctg\frac(\pi)(\sqrt(2n-1))$ we replace the arctangent with the fraction $\frac(\pi)(\sqrt(2n- 1))$. We get the following: $\frac(1)(\sqrt(n))\frac(\pi)(\sqrt(2n-1))$. We have already worked with such fractions before. Discarding the “extra” elements, we arrive at the fraction $\frac(1)(\sqrt(n)\cdot\sqrt(n))=\frac(1)(n^(\frac(1)(2)+\frac (1)(3)))=\frac(1)(n^(\frac(5)(6)))$. It is with the series $\sum\limits_(n=1)^(\infty)\frac(1)(n^\frac(5)(6))$ that we will compare the given series using . Since $\frac(5)(6)≤ 1$, then the series $\sum\limits_(n=1)^(\infty)\frac(1)(n^\frac(5)(6))$ diverges.

$$ \lim_(n\to\infty)\frac(\frac(1)(\sqrt(n))\arctg\frac(\pi)(\sqrt(2n-1)))(\frac(1) (n^\frac(5)(6)))=\left|\frac(0)(0)\right|=\left|\begin(aligned)&\frac(\pi)(\sqrt(2n- 1))\to 0;\\&\arctg\frac(\pi)(\sqrt(2n-1))\sim\frac(\pi)(\sqrt(2n-1)).\end(aligned) \right| =\lim_(n\to\infty)\frac(\frac(1)(\sqrt(n))\cdot\frac(\pi)(\sqrt(2n-1)))(\frac(1)( n^\frac(5)(6))) =\\=\pi\cdot\lim_(n\to\infty)\frac(\sqrt(n))(\sqrt(2n-1)) =\pi \cdot\lim_(n\to\infty)\frac(1)(\sqrt(2-\frac(1)(n)))=\pi\cdot\frac(1)(\sqrt(2-0) )=\frac(\pi)(\sqrt(2)). $$

Since $0<\frac{\pi}{\sqrt{2}}<\infty$, то ряды $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n}}\arctg\frac{\pi}{\sqrt{2n-1}}$ и $\sum\limits_{n=1}^{\infty}\frac{1}{n^\frac{5}{6}}$ сходятся либо расходятся одновременно. Так как ряд $\sum\limits_{n=1}^{\infty}\frac{1}{n^\frac{5}{6}}$ расходится, то одновременно с ним будет расходиться и ряд $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n}}\arctg\frac{\pi}{\sqrt{2n-1}}$.

I note that in in this case instead of the arctangent in the expression of the general term of the series there could be a sine, arcsine or tangent. The solution would remain the same.

Answer: the series diverges.

Example No. 10

Examine the series $\sum\limits_(n=1)^(\infty)\left(1-\cos\frac(7)(n)\right)$ for convergence.

Since the lower limit of summation is 1, the common term of the series is written under the sum sign: $u_n=1-\cos\frac(7)(n)$. Since for any value $x$ we have $-1≤\cos x≤ 1$, then $\cos\frac(7)(n)≤ 1$. Therefore, $1-\cos\frac(7)(n)≥ 0$, i.e. $u_n≥ 0$. We are dealing with a positive series.

If $n\to\infty$, then $\frac(7)(n)\to 0$. Therefore, $1-\cos\frac(7)(n)\sim \frac(\left(\frac(7)(n)\right)^2)(2)=\frac(49)(2n^2) $. Why is that? If we look at the table at the end of this document, we will see the formula $1-\cos x \sim \frac(x^2)(2)$ for $x\to 0$. We used this formula, only in our case $x=\frac(7)(n)$.

Let's replace the expression $1-\cos\frac(7)(n)$ with $\frac(49)(2n^2)$. Discarding the “extra” elements, we arrive at the fraction $\frac(1)(n^2)$. It is with the series $\sum\limits_(n=1)^(\infty)\frac(1)(n^2)$ that we will compare the given series using . Since $2 > 1$, the series $\sum\limits_(n=1)^(\infty)\frac(1)(n^2)$ converges.

$$ \lim_(n\to\infty)\frac(1-\cos\frac(7)(n))(\frac(1)(n^2))=\left|\frac(0)(0 )\right|= \left|\begin(aligned)&\frac(7)(n)\to 0;\\&1-\cos\frac(7)(n)\sim\frac(49)(2n^ 2).\end(aligned)\right| =\lim_(n\to\infty)\frac(\frac(49)(2n^2))(\frac(1)(n^2))=\frac(49)(2). $$

Since $0<\frac{49}{2}<\infty$, то ряды $\sum\limits_{n=1}^{\infty}\left(1-\cos\frac{7}{n}\right)$ и $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$ сходятся либо расходятся одновременно. Так как ряд $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$ сходится, то одновременно с ним будет сходиться и ряд $\sum\limits_{n=1}^{\infty}\left(1-\cos\frac{7}{n}\right)$.

Answer: the series converges.

Example No. 11

Investigate the convergence of the series $\sum\limits_(n=1)^(\infty)n\left(e^\frac(3)(n)-1\right)^2$.

Since the lower limit of summation is 1, the general term of the series is written under the sum sign: $u_n=n\left(e^\frac(3)(n)-1\right)^2$. Since both factors are positive, then $u_n >0$, i.e. we are dealing with a positive series.

If $n\to\infty$, then $\frac(3)(n)\to 0$. Therefore, $e^\frac(3)(n)-1\sim\frac(3)(n)$. The formula we used is located in the table at the end of this document: $e^x-1 \sim x$ at $x\to 0$. In our case, $x=\frac(3)(n)$.

Let us replace the expression $e^\frac(3)(n)-1$ with $\frac(3)(n)$, thereby obtaining $n\cdot\left(\frac(3)(n)\right)^ 2=\frac(9)(n)$. Removing the number, we arrive at the fraction $\frac(1)(n)$. It is with the harmonic series $\sum\limits_(n=1)^(\infty)\frac(1)(n)$ that we will compare the given series using . Let me remind you that the harmonic series diverges.

$$ \lim_(n\to\infty)\frac(n\left(e^\frac(3)(n)-1\right)^2)(\frac(1)(n))=\lim_( n\to\infty)\frac(\left(e^\frac(3)(n)-1\right)^2)(\frac(1)(n^2)) =\left|\frac(0 )(0)\right|=\left|\begin(aligned)&\frac(3)(n)\to 0;\\&e^\frac(3)(n)-1\sim\frac(3) (n).\end(aligned)\right| =\lim_(n\to\infty)\frac(\frac(9)(n^2))(\frac(1)(n^2))=9. $$

Since $0<9<\infty$, то одновременно с рядом $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ будет расходиться и ряд $\sum\limits_{n=1}^{\infty}n\left(e^\frac{3}{n}-1\right)^2$.

Answer: the series diverges.

Example No. 12

Investigate the convergence of the series $\sum\limits_(n=1)^(\infty)\ln\frac(n^3+7)(n^3+5)$.

Since the lower limit of summation is 1, the general term of the series is written under the sum sign: $u_n=\ln\frac(n^3+7)(n^3+5)$. Since for any value of $n$ we have $n^3+7 > n^3+5$, then $\frac(n^3+7)(n^3+5) > 1$. Therefore, $\ln\frac(n^3+7)(n^3+5) > 0$, i.e. $u_n > 0$. We are dealing with a positive series.

It is somewhat difficult to notice the equivalence that is needed in this case. Let's write the expression under the logarithm in a slightly different form:

$$ \ln\frac(n^3+7)(n^3+5)=\ln\frac(n^3+5+2)(n^3+5)=\ln\left(\frac( n^3+5)(n^3+5)+\frac(2)(n^3+5)\right)=\ln\left(1+\frac(2)(n^3+5)\ right). $$

Now the formula is visible: $\ln(1+x)\sim x$ for $x\to 0$. Since for $n\to\infty$ we have $\frac(2)(n^3+5)\to 0$, then $\ln\left(1+\frac(2)(n^3+5) \right)\sim\frac(2)(n^3+5)$.

Let's replace the expression $\ln\frac(n^3+7)(n^3+5)$ with $\frac(2)(n^3+5)$. Discarding the “extra” elements, we arrive at the fraction $\frac(1)(n^3)$. It is with the series $\sum\limits_(n=1)^(\infty)\frac(1)(n^3)$ that we will compare the given series using . Since $3 > 1$, the series $\sum\limits_(n=1)^(\infty)\frac(1)(n^3)$ converges.

$$ \lim_(n\to\infty)\frac(\ln\frac(n^3+7)(n^3+5))(\frac(1)(n^3))=\lim_(n \to\infty)\frac(\ln\left(1+\frac(2)(n^3+5)\right))(\frac(1)(n^3))=\left|\frac( 0)(0)\right|= \left|\begin(aligned)&\frac(2)(n^3+5)\to 0;\\&\ln\left(1+\frac(2)( n^3+5)\right)\sim\frac(2)(n^3+5).\end(aligned)\right|=\\ =\lim_(n\to\infty)\frac(\frac (2)(n^3+5))(\frac(1)(n^3)) =\lim_(n\to\infty)\frac(2n^3)(n^3+5)=\lim_ (n\to\infty)\frac(2)(1+\frac(5)(n^3))=\frac(2)(1+0)=2. $$

Since $0<2<\infty$, то одновременно с рядом $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}$ сходится и ряд $\sum\limits_{n=1}^{\infty}\ln\frac{n^3+7}{n^3+5}$.

Answer: the series converges.

Example No. 13

Explore the series $\sum\limits_(n=1)^(\infty)\frac(n^n)(7^n\cdot n$ на сходимость.!}

Since the lower limit of summation is 1, the common term of the series is written under the sum sign: $u_n=\frac(n^n)(7^n\cdot n$. Так как $u_n ≥ 0$, то заданный ряд является положительным.!}

Let a positive number series $ \sum_(n=1) ^\infty a_n $ be given. Let us formulate the necessary criterion for the convergence of a series:

1. If the series converges, then the limit of its common term is zero: $$ \lim _(n \to \infty) a_n = 0 $$
2. If the limit of the common term of the series is not equal to zero, then the series diverges: $$ \lim _(n \to \infty) a_n \neq 0 $$

Generalized harmonic series

This series is written as follows: $ \sum_(n=1) ^\infty \frac(1)(n^p) $. Moreover, depending on $p$, the series converges or diverges:

1. If $ p = 1 $, then the series $ \sum_(n=1) ^\infty \frac(1)(n) $ diverges and is called harmonic, despite the fact that the common term $ a_n = \frac(1)( n) \to 0 $. Why is that? The remark said that the necessary criterion does not give an answer about the convergence, but only about the divergence of the series. Therefore, if we apply a sufficient criterion, such as the integral Cauchy criterion, it becomes clear that the series diverges!
2. If $ p \leqslant 1 $, then the series diverges. Example, $ \sum_(n=1) ^\infty \frac(1)(\sqrt(n)) $, in which $ p = \frac(1)(2) $
3. If $p > 1$, then the series converges. Example, $ \sum_(n=1) ^\infty \frac(1)(\sqrt(n^3)) $, in which $ p = \frac(3)(2) > 1 $

Examples of solutions

Example 1

Prove the divergence of the series $ \sum_(n=1) ^\infty \frac(n)(6n+1) $

Solution

The series is positive, we write down the common term: $$ a_n = \frac(n)(6n+1) $$ We calculate the limit at $ n \to \infty $: $$ \lim _(n \to \infty) \frac(n)(6n+1) = \frac(\infty)(\infty) = $$ We take $ n $ out of brackets in the denominator, and then perform a reduction on it: $$ = \lim_(n \to \infty) \frac(n)(n(6+\frac(1)(n))) = \lim_(n \to \infty) \frac(1)(6 + \frac(1)(n)) = \frac(1)(6) $$ Since we found that $ \lim_(n\to \infty) a_n = \frac(1)(6) \neq 0 $, then the necessary Cauchy test is not satisfied and the series therefore diverges. If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner!

Answer

The series diverges

Application

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Rows for dummies. Examples of solutions

I welcome all survivors to the second year! In this lesson, or rather, in a series of lessons, we will learn how to manage rows. The topic is not very complicated, but mastering it will require knowledge from the first year, in particular, you need to understand what is a limit, and be able to find the simplest limits. However, it’s okay, as I explain, I will provide relevant links to the necessary lessons. To some readers, the topic of mathematical series, solution methods, signs, theorems may seem peculiar, and even pretentious, absurd. In this case, you don’t need to be too “loaded”; we accept the facts as they are and simply learn to solve typical, common tasks.

1) Rows for dummies, and for samovars immediately content :)

For super-fast preparation on the topic There is an express course in pdf format, with the help of which you can really “raise” your practice literally in a day.

The concept of a number series

IN general view number series can be written like this: .
Here:
– mathematical sum icon;
– common term of the series(remember this simple term);
– “counter” variable. The notation means that summation is carried out from 1 to “plus infinity”, that is, first we have , then , then , and so on - to infinity. Instead of a variable, a variable or is sometimes used. Summation does not necessarily start from one; in some cases it can start from zero, from two, or from any natural number.

In accordance with the “counter” variable, any series can be expanded:
- and so on, ad infinitum.

Components - This NUMBERS which are called members row. If they are all non-negative (greater than or equal to zero), then such a series is called positive number series.

Example 1

This, by the way, is already a “combat” task - in practice, quite often it is necessary to write down several terms of a series.

First, then:
Then, then:
Then, then:

The process can be continued indefinitely, but according to the condition it was required to write the first three terms of the series, so we write down the answer:

Please note the fundamental difference from number sequence,
in which the terms are not summed up, but are considered as such.

Example 2

Write down the first three terms of the series

This is an example for independent decision, answer at the end of the lesson

Even for a series that is complex at first glance, it is not difficult to describe it in expanded form:

Example 3

Write down the first three terms of the series

In fact, the task is performed orally: mentally substitute into the common term of the series first, then and. Eventually:

We leave the answer as follows: It is better not to simplify the resulting series terms, that is do not perform actions: , , . Why? The answer is in the form it is much easier and more convenient for the teacher to check.

Occasionally occurs reverse reference

Example 4

There is no clear solution algorithm here, you just need to see the pattern.
In this case:

To check, the resulting series can be “written back” in expanded form.

Here's an example that's a little more complicated to solve on your own:

Example 5

Write down the sum in collapsed form with the common term of the series

Perform a check by again writing the series in expanded form

Convergence of number series

One of key tasks topic is study of series for convergence. In this case, two cases are possible:

1) Rowdiverges. It means that infinite sum equal to infinity: or sums in general does not exist, as, for example, in the series
(here, by the way, is an example of a series with negative terms). Nice sample divergent number series was encountered at the beginning of the lesson: . Here it is quite obvious that each next member of the series is greater than the previous one, therefore and, therefore, the series diverges. An even more trivial example: .

2) Rowconverges. This means that an infinite sum is equal to some finite number: . Please: – this series converges and its sum is zero. As a more meaningful example, we can cite infinitely decreasing geometric progression, known to us since school: . The sum of terms is infinitely decreasing geometric progression is calculated using the formula: , where is the first term of the progression, and is its base, which is usually written in the form correct fractions In this case: , . Thus: A finite number is obtained, which means the series converges, which is what needed to be proved.

However, in the vast majority of cases find the sum of the series is not so simple, and therefore in practice, to study the convergence of a series, special signs that have been proven theoretically are used.

There are several signs of series convergence: necessary test for the convergence of a series, comparison tests, D'Alembert's test, Cauchy's tests, Leibniz's sign and some other signs. When to use which sign? It depends on the common member of the series, figuratively speaking, on the “filling” of the series. And very soon we will sort everything out.

! To further learn the lesson, you must understand well what is a limit and it is good to be able to reveal the uncertainty of a type. To review or study the material, please refer to the article Limits. Examples of solutions.

A necessary sign of convergence of a series

If a series converges, then its common term tends to zero: .

Reverse to general case false, i.e., if , then the series can either converge or diverge. And therefore this sign is used to justify divergences row:

If the common term of the series does not tend to zero, then the series diverges

Or in short: if , then the series diverges. In particular, a situation is possible where the limit does not exist at all, as, for example, limit. So they immediately justified the divergence of one series :)

But much more often, the limit of a divergent series is equal to infinity, and instead of “x” it acts as a “dynamic” variable. Let's refresh our knowledge: limits with “x” are called limits of functions, and limits with the variable “en” are called limits of numerical sequences. The obvious difference is that the variable "en" takes discrete (discontinuous) natural values: 1, 2, 3, etc. But this fact has little effect on methods for solving limits and methods for disclosing uncertainties.

Let us prove that the series from the first example diverges.
Common member of the series:

Conclusion: row diverges

The necessary feature is often used in real practical tasks:

Example 6

We have polynomials in the numerator and denominator. The one who carefully read and comprehended the method of disclosing uncertainty in the article Limits. Examples of solutions, I probably caught that when the highest powers of the numerator and denominator equal, then the limit is finite number .

Divide the numerator and denominator by

Series under study diverges, since the necessary criterion for the convergence of the series is not fulfilled.

Example 7

Examine the series for convergence

This is an example for you to solve on your own. Complete solution and the answer at the end of the lesson

So, when we are given ANY number series, Firstly we check (mentally or on a draft): does its common term tend to zero? If it doesn’t, we formulate a solution based on examples No. 6, 7 and give an answer that the series diverges.

What types of apparently divergent series have we considered? It is immediately clear that series like or diverge. The series from examples No. 6, 7 also diverge: when the numerator and denominator contain polynomials, and the leading power of the numerator is greater than or equal to the leading power of the denominator. In all these cases, when solving and preparing examples, we use the necessary sign of convergence of the series.

Why is the sign called necessary? Understand in the most natural way: in order for a series to converge, necessary, so that its common term tends to zero. And everything would be great, but there’s more not enough. In other words, if the common term of a series tends to zero, THIS DOES NOT MEAN that the series converges– it can both converge and diverge!

Meet:

This series is called harmonic series. Please remember! Among the number series, he is a prima ballerina. More precisely, a ballerina =)

It's easy to see that , BUT. In theory mathematical analysis it has been proven that harmonic series diverges.

You should also remember the concept of a generalized harmonic series:

1) This row diverges at . For example, the series , , diverge.
2) This row converges at . For example, the series , , , converge. I emphasize once again that in almost all practical tasks it is not at all important to us what the sum of, for example, the series is equal to, the very fact of its convergence is important.

These are elementary facts from the theory of series that have already been proven, and when solving any practical example one can safely refer, for example, to the divergence of a series or the convergence of a series.

In general, the material in question is very similar to study of improper integrals, and it will be easier for those who have studied this topic. Well, for those who haven’t studied it, it’s doubly easier :)

So, what to do if the common term of the series TENDS to zero? In such cases, to solve examples you need to use others, sufficient signs of convergence/divergence:

Comparison criteria for positive number series

I draw your attention, that here we are talking only about positive number series (with non-negative terms).

There are two signs of comparison, one of them I will simply call a sign of comparison, another - limit of comparison.

Let's first consider comparison sign, or rather, the first part of it:

Consider two positive number series and . If known, that the series – converges, and, starting from some number, the inequality is satisfied, then the series also converges.

In other words: From the convergence of the series with larger terms follows the convergence of the series with smaller terms. In practice, the inequality often holds for all values:

Example 8

Examine the series for convergence

First, let's check(mentally or in draft) execution:
, which means it was not possible to “get off with little blood.”

We look into the “pack” of the generalized harmonic series and, focusing on the highest degree, we find a similar series: It is known from theory that it converges.

For all natural numbers, the obvious inequality holds:

and the larger denominators correspond to smaller fractions:
, which means, based on the comparison criterion, the series under study converges together with next to .

If you have any doubts, you can always describe the inequality in detail! Let us write down the constructed inequality for several numbers “en”:
If , then
If , then
If , then
If , then
….
and now it is absolutely clear that inequality fulfilled for all natural numbers “en”.

Let's analyze the comparison criterion and the solved example from an informal point of view. Still, why does the series converge? Here's why. If a series converges, then it has some final amount: . And since all members of the series less corresponding terms of the series, then it is clear that the sum of the series cannot be more number, and even more so, cannot be equal to infinity!

Similarly, we can prove the convergence of “similar” series: , , etc.

! note, that in all cases we have “pluses” in the denominators. The presence of at least one minus can seriously complicate the use of the product in question. comparison sign. For example, if a series is compared in the same way with a convergent series (write out several inequalities for the first terms), then the condition will not be satisfied at all! Here you can dodge and select another convergent series for comparison, for example, but this will entail unnecessary reservations and other unnecessary difficulties. Therefore, to prove the convergence of a series it is much easier to use limit of comparison(see next paragraph).

Example 9

Examine the series for convergence

And in this example, I suggest you consider for yourself second part of the comparison attribute:

If known, that the series – diverges, and starting from some number (often from the very first), the inequality is satisfied, then the series also diverges.

In other words: From the divergence of a series with smaller terms follows the divergence of a series with larger terms.

What should be done?
It is necessary to compare the series under study with a divergent harmonic series. For better understanding construct several specific inequalities and make sure that the inequality is fair.

The solution and sample design are at the end of the lesson.

As already noted, in practice, the comparison criterion just discussed is rarely used. The real workhorse of number series is limit of comparison, and in terms of frequency of use it can only compete with d'Alembert's sign.

Limit test for comparing numerical positive series

Consider two positive number series and . If the limit of the ratio of the common terms of these series is equal to finite non-zero number: , then both series converge or diverge simultaneously.

When is the limiting criterion used? The limiting criterion for comparison is used when the “filling” of the series is polynomials. Either one polynomial in the denominator, or polynomials in both the numerator and denominator. Optionally, polynomials can be located under the roots.

Let's deal with the row for which the previous comparison sign has stalled.

Example 10

Examine the series for convergence

Let's compare this series with a convergent series. We use the limiting criterion for comparison. It is known that the series converges. If we can show that equals finite, non-zero number, it will be proven that the series also converges.

A finite non-zero number is obtained, which means the series under study is converges together with next to .

Why was the series chosen for comparison? If we had chosen any other series from the “cage” of the generalized harmonic series, then we would not have succeeded in the limit finite, non-zero numbers (you can experiment).

Note: when we use the limiting comparison criterion, doesn't matter, in what order to compose the relation of common members, in the example considered, the relation could be compiled the other way around: - this would not change the essence of the matter.