Converge?
$\displaystyle
\sum\limits_{n = 1}^\infty {\frac{{\sin (n)}}
{n}}
$
thanks
Yes. It can be proven by Dirichlet's test that $\displaystyle \sum_{n=1}^{\infty}\frac{\sin n}{n}$ and $\displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n}$ converge.
It is more interesting to sum this series.
Remember that $\displaystyle 2i\sin (x) = \sinh (ix)$.
Therefore,
$\displaystyle \sum_{n=1}^{\infty} \frac{\sin (n)}{n} = \frac{1}{2i} \left( \sum_{n=1}^{\infty} \frac{e^{in}}{n} - \sum_{n=1}^{\infty} \frac{e^{-in}}{n} \right)$
Now use the formula,
$\displaystyle \sum_{n=1}^{\infty} \frac{z^n}{n} = - \log (1-z)$ for $\displaystyle |z| < 1$
Therefore, we get,
$\displaystyle \frac{1}{2i}\left( \log (1 - e^{-i}) - \log (1 - e^{i}) \right) = \sin 1$
I hope I did not make any mistakes.
Maybe we can solve delimiting?
I always find other serie minor, but this coverge or a major, but this diverge...
Too I try whit quotien rule o the integral, but never obey the hypothesis
Moreover I am in first year of university, so I dont know powerful tool...
anyways...thank you ThePerfectHacker