Investigate the convergence of a number series of examples. Rows for teapots

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

If the series converges, then the limit of its common term is equal to zero: $$ \lim _(n \to \infty) a_n = 0 $$
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:

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 test, such as the integral Cauchy test, then it becomes clear that the series diverges!
If $ p \leqslant 1 $, then the series diverges. Example, $ \sum_(n=1) ^\infty \frac(1)(\sqrt(n)) $, where $ p = \frac(1)(2) $
If $ p > 1 $, then the series converges. Example, $ \sum_(n=1) ^\infty \frac(1)(\sqrt(n^3)) $, where $ p = \frac(3)(2) > 1 $

Solution examples

Example 1
Prove the divergence of the series $ \sum_(n=1) ^\infty \frac(n)(6n+1) $
Decision
The series is positive, we write down the common term: $$ a_n = \frac(n)(6n+1) $$ Calculate the limit at $ n \to \infty $: $$ \lim _(n \to \infty) \frac(n)(6n+1) = \frac(\infty)(\infty) = $$ Bracket $ n $ in the denominator, and then reduce 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 have obtained that $ \lim_(n\to \infty) a_n = \frac(1)(6) \neq 0 $, the necessary Cauchy criterion is not satisfied and the series therefore diverges. If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the teacher in a timely manner!
Answer
The series diverges

Example #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 summation limit is 1, the common 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 $\frac(1)(\sqrt(n))> 0$, that's clear, but what about the arc tangent? There is nothing complicated with arctangues: 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 test to study the question of the convergence of this series.

First, let's choose a series with which we will compare. If $n\to\infty$ then $\frac(\pi)(\sqrt(2n-1))\to 0$. Hence $\arctg\frac(\pi)(\sqrt(2n-1))\sim\frac(\pi)(\sqrt(2n-1))$. 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 $\frac(1)(\sqrt(n))\arctg\frac(\pi)(\sqrt(2n-1))$ the arc tangent by 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 $\frac(1)(\sqrt(n)\cdot\sqrt(n))=\frac(1)(n^(\frac(1)(2)+\frac (1)(3)))=\frac(1)(n^(\frac(5)(6)))$. 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 this case, instead of the arc tangent in the expression of the common term of the series, there could be a sine, arc sine or tangent. The solution would remain the same.

Answer: the series diverges.

Example #10

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

Since the lower summation limit is equal to 1, the common term of the series is written under the sum sign: $u_n=1-\cos\frac(7)(n)$. Since for any value of $x$ 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 $\frac(7)(n)\to 0$. Hence $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 us 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 #11

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

Since the lower summation limit is 1, the common 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 $\frac(3)(n)\to 0$. Hence $e^\frac(3)(n)-1\sim\frac(3)(n)$. The formula we used is in the table at the end of this document: $e^x-1 \sim x$ for $x\to 0$. In our case $x=\frac(3)(n)$.

Let us replace the expression $e^\frac(3)(n)-1$ by $\frac(3)(n)$, thus obtaining $n\cdot\left(\frac(3)(n)\right)^ 2=\frac(9)(n)$. Discarding 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 #12

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

Since the lower summation limit is 1, the common 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.

Noticing the equivalence that is needed in this case is somewhat difficult. 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 us 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 #13

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

Since the lower summation limit 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.!}

Appendix

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