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At first sight I answered that this series is divergent to my teacher, but just after a moment I realized that this is a tricky series with the term alternating, the difficult part of this question is that the terms of this series does not alternate one after another.

Liding Yao
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Ketan Choudhary
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    Try Dirichlet's test. If you are familiar with $\sin(n)$ being the imaginary part of $e^{in}$, it is much easier to show the sequence $\sin(n)$ has bounded partial sums. – Mark May 01 '24 at 00:30
  • @Mark ok thanks.. – Ketan Choudhary May 01 '24 at 00:32
  • I do not have a proof so far, but at least mathematica states that the series converges, with the limit being $1/2 * (\pi-1)$ – HyperPro May 01 '24 at 00:54

2 Answers2

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The trick is called summation by parts:

How do you show $\int_1^\infty\frac{\sin x}xdx$ converges in the sense of improper Riemann integral? IbP gives you $\int_1^\infty\frac{\sin x}xdx=-\frac{\cos x}x|_{x=1}^\infty-\int_1^\infty\frac{\cos x}{x^2}dx$, the right hand side of which converges.

For series we have $\sum_{n=1}^\infty (A_{n+1}-A_n)B_n=A_nB_n|_{n=1}^\infty-\sum_{n=1}^\infty A_{n+1}(B_{n+1}-B_n)$. Choose $A_n=\sum_{k=1}^n\sin k$ and $B_n=\frac1n$ this is what you need.

Finally notice that $A_n=\sum_{k=1}^n\sin k$ is a bounded sequence because $$\sum_{k=1}^{n} \sin k= \frac{\sin1+\sin n-\sin(n+1)}{2(1-\cos 1)}.$$ Combining with $B_{n+1}-B_n=\frac1{n(n+1)}=O(n^{-2})$, this concludes that $\sum_{n=1}^\infty |A_{n+1}(B_{n+1}-B_n)|<\infty$.

Liding Yao
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As $-\frac1n\le\frac{sin(n)}{n}\le\frac1n$, so when n goes to infinity, $0\le\frac{sin(n)}{n}\le0$, so $a_n$ converges.

Mrexcel
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    This argues that the sequence converges, not the series. – Randall May 01 '24 at 00:36
  • Probably what is meant is $\sum \frac{\sin(n)}{n}$, not just the sequence $\sin(n)/n$. – Ian May 01 '24 at 00:36
  • @Ian Even if so, it is still wrong. Comparison tests can be only used for series on non negative terms. And the harmonic series is divergent anyway. – Mark May 01 '24 at 09:53
  • More precisely I meant "probably what is meant in the OP is the series, not the sequence, so saying the sequence goes to zero doesn't allow us to conclude anything about the series". I was not replying to Randall. – Ian May 01 '24 at 11:15
  • @Ian Oh, ok then. – Mark May 01 '24 at 13:06