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.
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 $N$.
$$ \sum_{n=1}^N b_n=\Im\sum_{n=1}^N (-e^i)^n=\Im\left(\frac{-e^i(1-(-e^{Ni}))}{1+e^i}\right)$$
$$\left|\sum_{n=1}^N b_n\right|=\left|\Im\left(\frac{-e^i(1-(-e^{Ni}))}{1+e^i}\right) \right|\le \left|\frac{-e^i(1-(-e^{Ni}))}{1+e^i} \right|\le\frac{2}{1}$$
My 2 cents: $$\left|\sum\limits_{k=1}^{n}(-1)^k \sin k \right|= \frac{1}{\left|\cos\frac{1}{2}\right|}\cdot\left|\sum\limits_{k=1}^{n}(-1)^k \sin k \cos\frac{1}{2} \right|=\\ =\frac{1}{2\left|\cos\frac{1}{2}\right|}\cdot\left|\sum\limits_{k=1}^{n}(-1)^k\left[\sin\frac{2k+1}{2} +\sin\frac{2k-1}{2}\right] \right|=\\ =\frac{1}{2\left|\cos\frac{1}{2}\right|}\cdot \left|-\sin\frac{1}{2}+ (-1)^n\sin\frac{2n-1}{2}\right|\leqslant \frac{1}{\left|\cos\frac{1}{2}\right|}$$
For any $\lambda\in(-1,1)$ we have
$$ \sum_{n\geq 1}\frac{(-\lambda e^i)^n}{n} = -\log(1+\lambda e^i) $$
so by taking the imaginary part
$$ \sum_{n\geq 1}\frac{(-1)^n\lambda^n \sin(n)}{n} = -\text{arg}(1+\lambda e^i)=-\arctan\left(\frac{\lambda\sin(1)}{1+\lambda\cos(1)}\right). $$
By considering the limit of both sides as $\lambda$ approaches $1^-$ we get
$$ \boxed{\sum_{n\geq 1}\frac{(-1)^n\sin(n)}{n} = \color{blue}{-\frac{1}{2}}}.$$
As an alternative, let $f(x)$ be the $2\pi$-periodic extension of the function which equals $-\frac{x}{2}$ on $(-\pi,\pi)$.
The Fourier series of $f(x)$ is given by
$$ \sum_{n\geq 1}\frac{(-1)^n \sin(nx)}{n} $$
and the evaluation at $x=1$ returns $-\frac{1}{2}$ since the Fourier series is uniformly convergent to $f(x)$ over any compact subset of $\mathbb{R}\setminus\pi\mathbb{Z}$.
