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I need to prove that $\sum \limits _{n=1}^{\infty}\frac{\sin (n)}{n}$ convergent only by Cauchy's test.

Cauchy's test for sequences convergence :

The $\sum \limits _{n=1}^{\infty}a_n$ convergent $\iff$ $\forall \varepsilon >0$ $ \exists$ $ n_0\in \mathbb{N}$ such that $\forall $ $n_0 \le n$ , $\forall$ $p\in \mathbb{N}$ the following holds $\mid S_{n+p}-S_n\mid = \mid a_{n+1}+\dots+a_{n+p}\mid<\varepsilon$

My Attempt:

Let $\varepsilon>0$,$\ p\in\mathbb{N}$. We define $n_0=\frac{p}{\varepsilon}$, and since $-1\le \sin(x)\le 1$ therefore, for each $n\le n_0$: $$\mid \frac{\sin (n+1)}{n+1}+\frac{\sin (n+2)}{n+2}+...+\frac{\sin (n+p)}{n+p} \mid \ \le \ \mid \frac{\sin (n+1)}{n+1}\mid + \mid \frac{\sin (n+2)}{n+2}\mid+...+\mid \frac{\sin (n+p)}{n+p} \mid \le \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+p}\le \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}=\frac{p}{n}<\varepsilon$$

Is my proof correct? if not so how can I prove it only by Cauchy's test as I said earlier, without any further test - such as Dirichlet, Abel...

Johan_Marsan
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2 Answers2

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There have been a couple of requests in the comments to show a way forward after I suggested applying summation by parts. To that end, we proceed.

Since we asked to prove that the series $\sum_{n=1}^\infty \frac{\sin(n)}{n}$ converges, we begin with the partial sum $S_N$ defined by

$$S_N=\sum_{n=1}^N \frac{\sin(n)}{n}$$

Letting $T_n=\sum_{m=1}^n \sin(m)$ and summing by parts, we have for $N>M+1$

$$\begin{align} \left|S_N-S_M\right|&=\left|\frac{T_N}{N+1}-\frac{T_M}{M+1}+\sum_{n=M+1}^N \left(\frac1n-\frac1{n+1}\right)T_n\right|\\\\ &\le \max_n (|T_n|)\left(\frac1{N+1}+\frac1{M+1}+\frac1{M+1}-\frac1{N+1}\right) \end{align}$$

I showed in THIS ANSWER that $\max_n |T_n|\le \frac12(\csc(1/2)+\cot(1/2))\le 2$.

Using this bound for $|T_n|$, we can assert that for any given $\varepsilon >0$, $|S_N-S_M|<\varepsilon$ whenever $N>M+1>\frac{4}{\varepsilon}$.

From Cauchy's criterion, we conclude that $S_N$ converges. And we are done!

Mark Viola
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    First of all , Thanks a lot for your solution , by the way I hardly can understand the $|S_N-S_M| = |\frac{\sin(N)}{N+1}-....$ this line . how you get it from the line before I will appreciate that if you will explain in a few words that will make me understand each line in the proof. I think that in the same line I asked you for .. the $|S_N-S_M| = |\frac{\sin(N)}{N+1}-....$ not should be $|S_N-S_M| = |\frac{\sin(N)}{N}-\frac{\sin(M+1)}{M+1}... ?$ – ATB Mar 23 '21 at 23:58
  • @ATB You're welcome. My pleasure. Please review summation by parts to gain an understanding of the posted solution. While it's possible that there is an error in the bookkeeping, I believe all is correct as is. When I have more time, I'll review and revert. Otherwise, try going through application of summation by parts on your own and see if you can verify the details. – Mark Viola Mar 24 '21 at 00:36
  • @ATB There was a bookkeeping error, which I've edited. – Mark Viola Mar 24 '21 at 03:58
  • I think the author of this question in this post mean that you solve without Abel's theorem or Abel transformation, just only by the Cauchy's criterion. – ATB Mar 24 '21 at 22:42
  • @ATB Summation by parts is not anything unusual. It's just arithmetic. So, I don't believe that the OP meant that one is prohibited from using any arithmetic whatsoever. And inasmuch as the OP accepted the answer, your supposition is rather dubious at best. – Mark Viola Mar 24 '21 at 22:56
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    I have accepted your answer because it seems to be right, however after thinking, I don't think I have learned the method you have used in your answer - "summation by parts". How can I show it by easy algebra moves? I will be happy to see how you will prove it with only algebra. Moreover, the inequality of $\sum_{n=1}^{\infty} \sin(n) \le \frac{2}{5}$, I don't think I have learned that either, or proved... is there an alternative for your answer? - but of course with my conditions mentioned in my question, if so, it will be great if you can add it to your answer. Thanks! – Johan_Marsan Mar 25 '21 at 09:55
  • Hi Johan. HERE is a link which discusses summation by parts. It is analogous to integration by parts, which you have likely learned. I added a link to an answer I posted that provides an upper bound estimate for $\left|\sum_{m=1}^n \sin(m)\right|$. In fact, it is bounded by $2$. – Mark Viola Mar 25 '21 at 14:46
  • It is so complicated for me, because I am a beginner with the sequences topic. Can you show it in the easiest way by Cauchy? because summation by parts and the bounded limit of $| T_n |$ I am not allowed to use. – Johan_Marsan Mar 25 '21 at 15:48
  • Johan, if you understand the proof of Cauchy's criterion, and understand integration by parts, then surely summation by parts and the way I developed a bound for $|T_n|$ are well within your grasp. I don't know a simpler way to prove convergence of $S_n$ – Mark Viola Mar 25 '21 at 16:45
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Your proof is wrong because you have $\sin(n)$ in the numerators where you should have $\sin(n+p)$.

marty cohen
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