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Determine the convergence of the sequence defined by $\sum_{n \in \mathbb{N}}a_n$ such as $$a_1=1,\quad a_{n+1}=\frac{2+\cos(n)}{\sqrt{n}}a_n$$

By the test of the general term, we have $$\frac{1}{\sqrt{n}} \le \frac{2 + \cos{(n)}}{\sqrt{n}} \le \frac{3}{\sqrt{n}} $$ So, by the theorem of the sandwich, $$ \lim_{n \rightarrow \infty}\frac{1}{\sqrt{n}} =\lim_{n \rightarrow \infty}\frac{2 + \cos{(n)}}{\sqrt{n}} =\lim_{n \rightarrow \infty}\frac{3}{\sqrt{n}} $$ We get, $\lim_{n \rightarrow \infty}\frac{2+\cos(n)}{\sqrt{n}}a_n=0$.

How to a continue my proof?

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  • $\begingroup$ How do you know, that $ (2+\cos(n))a_n $ is bounded? $\endgroup$
    – iiivooo
    Oct 20, 2015 at 0:11
  • $\begingroup$ @iiivooo I think we don't have to bound $(2+\cos{(n)})a_n$, if we bound $2+\cos{(n)}$, Q.E.D. $\endgroup$
    – hlapointe
    Oct 24, 2015 at 1:57

3 Answers 3

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If we use the ratio test, $$ \left| \frac{a_{n+1}}{a_n} \right| = \left|\frac{2+\cos{(n)}}{\sqrt n} \right| \leq \frac{3}{\sqrt{n}} \to 0. $$ So the series converges absolutely.

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  • $\begingroup$ Nice. Can we also prove it by findind the bouderies or the sequence? $\endgroup$
    – hlapointe
    Oct 20, 2015 at 0:29
  • $\begingroup$ Why can we say $\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{2+\cos n}{\sqrt n}\right|$. Can you show me the intermediate steps? $\endgroup$
    – hlapointe
    Oct 20, 2015 at 0:34
  • $\begingroup$ Why can we suppose that $a_n = a_1 = 1$ for the test of the ratio? $\endgroup$
    – hlapointe
    Oct 20, 2015 at 0:39
  • $\begingroup$ Are you sure that $\left\| \frac{2+\cos{n}}{\sqrt{n}} \right\| \le \frac{2}{\sqrt{n}}$. Because $2+\cos{n} \le 3$. $\endgroup$
    – hlapointe
    Oct 20, 2015 at 0:43
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    $\begingroup$ If we had $a_n=1$ then the series wouldn't converge; it wouldn't satisfy your recursion formula either. About the "intermediate steps": since $$a_{n+1}=\frac{2+\cos(n)}{\sqrt{n}}a_n,$$ you divide both sides by $a_n$ and you get $$\frac{a_{n+1}}{a_n}=\frac{2+\cos(n)}{\sqrt{n}}.$$ And you are right, it should have been a $3$. $\endgroup$
    – Martin Argerami
    Oct 20, 2015 at 0:44
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The problem here is that $\lim_{n\to\infty}(2+\cos n)$. The idea you need is that $1\le 2+\cos n\le 3$. So, $$\frac 1{\sqrt n}\le\frac{2+\cos n}{\sqrt n}\le\frac 3{\sqrt n}.$$

Now, use the squeeze theorem.

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    $\begingroup$ Ok. I understant my mistake. But that doesn't proof the convergence of the sequence? $\endgroup$
    – hlapointe
    Oct 20, 2015 at 0:14
  • $\begingroup$ Not without using the fact that $2+\cos n$ is bounded. For example, $\lim (n+1)/\sqrt n=\infty$. $\endgroup$
    – Tim Raczkowski
    Oct 20, 2015 at 0:16
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For $n\geq 16 $ we have $|a_{n+1}|=(|2+\cos n|/\sqrt n)|a_n|\leq(3/4)|a_n|$ so for $n\geq 16$ we have $|a_{n+1}|\leq (3/4)^{n-15}|a_{16}|.$

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  • $\begingroup$ it would if you knew anything about comparison tests and absolute convergence of geometric series. $\endgroup$
    – DanielWainfleet
    Oct 20, 2015 at 17:08

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