How shall I prove the convergence of series $\sum\limits_{n=1}^{+\infty}\frac{(-1)^n\sin (n)}{n} ~?$
I'm trying to use Dirichlet test since $\frac 1 n$ is monotonically decreasing to $0$, but I don't know how to prove that $\sum_{n=1}^\infty (-1)^n\sin (n)$ is bounded.
Can someone help me, please? Thanks a lot.
Best Answer
Let $a_n=\frac{1}n$, so $a_n$ is monotonic decreasing and $\lim a_n=0$. Let $b_n=(-1)^n \sin(n)$ and we want to show $\left|\sum_{n=1}^N b_n\right|$ is bounded for any $N$.
$$ \sum_{n=1}^N b_n=\Im\sum_{n=1}^N (-e^i)^n=\Im\left(\frac{-e^i(1-(-e^{Ni}))}{1+e^i}\right)$$
$$\left|\sum_{n=1}^N b_n\right|=\left|\Im\left(\frac{-e^i(1-(-e^{Ni}))}{1+e^i}\right) \right|\le \left|\frac{-e^i(1-(-e^{Ni}))}{1+e^i} \right|\le\frac{2}{1}$$