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…
Best Answer
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!