Prove the sequence $a_n=\sum_{k=1}^n \frac {\sin(k)}{k^2}$ converges

Question

I'm stuck at this question.

I tried proving that the sequence converges through proving the sequence is Cauchy but I'm stuck there too.

My try : Let $\epsilon>0$ and $n\ge m$.

$$|a_n-a_m|=\left|\sum_{k=1}^n\frac{\sin(k)}{k^2}-\sum_{k=1}^m\frac{\sin(k)}{k^2}\right|\\ =\left|\sum_{k=m+1}^{n}\frac{\sin(k)}{k^2}\right|$$

And exactly there I'm stuck.

How to prove it ? Thanks.

Answer 1

Note that$$\left|\sum_{k=m+1}^n\frac{\sin(k)}{k^2}\right|\leqslant\sum_{k=m+1}^n\frac1{k^2}$$and that the sequence $\left(\sum_{k=1}^n\frac1{k^2}\right)_{n\in\Bbb N}$ is a Cauchy sequence, since the series $\sum_{k=1}^\infty\frac1{k^2}$ converges.