Calculating $\sum\limits_{n=1}^{\infty}\frac{(-1)^n \sin(n)}{n}$ I'm trying to evaluate the following sum:
$$\displaystyle\sum\limits_{n=1}^{\infty}\frac{(-1)^n \sin(n)}{n}$$
But I'm having trouble getting anywhere. Wolphram Alpha indicated some stragety that uses complex logarithms in order to find the answer, which seems to be $-0.5$.
Anyone know a good strategy here?
 A: Running with my initial comment, it turns out there is a way to make this strategy work.
Consider the function $s(x) = x/\pi$ when $x \in  (- \pi, \pi)$, and extend this function to be periodic of period $2 \pi$ on the whole real line. This function is piecewise smooth, and so has a Fourier series, which is calculated in this wikipedia link. The resulting series is $$s(x) = \frac{2}{\pi}\sum_{n=1}^\infty \frac{(-1)^{n+1}\sin(nx)}{n} = -\frac{2}{\pi}\sum_{n=1}^\infty \frac{(-1)^n\sin(nx)}{n}$$
Which is a number times your sum. Denoting this sum as $S(x)$, and evaluating $s(x)$ at $1$, we get $\frac{1}{\pi}$, so $$s(1) = -\frac{2}{\pi}S(1) = \frac{1}{\pi}$$
This implies the answer Wolfram Alpha gave!
A: Using  
$$\sin(x) = \Im( \exp(i x))\tag{1}$$ 
and for $|z|\le 1, z\ne 1$
$$\sum_{n=1}^{\infty}(-1)^n z^n/n=-\log(1+z)\tag{2}$$ 
we can write
$$
\begin{align}
&s = \sum_{n=1}^{\infty} (-1)^n \sin(n) / n \\
&= \Im\sum_{n=1}^{\infty} (-1)^n \exp(i\; n) / n \\
&= - \Im \log (1+\exp(i))\\
& = - \frac{1} {2i}\left( \log (1+\exp(i))- \log (1+\exp(-i))\right)\\
&= - \frac{1} {2i} \log \left(\frac{ 1+\exp(i)}{1+\exp(-i)}\right)\\
&= - \frac{1} {2i} \log \left(\exp(i)\frac{ 1+\exp(i)}{\exp(i)+1}\right)\\
&= - \frac{1} {2i} \log \left(\exp(i)\right)\\
&= - \frac{1} {2i} i = -\frac{1}{2}
\end{align}
$$
Additional question:    
Calculate the explicit expresion for the sum with $\sin$ replaced by $\cos$
A: 
I thought it might be instructive to present an approach that relies on straightforward, first year calculus methodologies.  To that end we now proceed.


Note that we can write
$$\begin{align}
\sum_{n=1}^N \frac{(-1)^n\sin(n)}{n}&=\int_0^1 \sum_{n=1}^N (-1)^n\cos(nx)\,dx\\\\
&=\int_0^1 \frac12 \left((-1)^N\cos((N+1)x)+(-1)^N\tan(x/2) \sin((N+1)x)-1\right) \,dx\tag1\\\\
&=-\frac12 +\frac12(-1)^N\,\frac{\sin(N+1)}{N+1}+\frac12(-1)^N\int_0^1 \tan(x/2)\sin((N+1)x)\,dx\tag2\\\\
&=-\frac12 +\frac12(-1)^N\,\frac{\sin(N+1)}{N+1}\\\\&+\frac12(-1)^N\left(-\frac{\tan(1/2)\cos((N+1))}{N+1}+\frac{1}{2(N+1)}\int_0^1 \sec^2(x/2)\cos((N+1)x)\,dx\right)\tag3
\end{align}$$

In arriving at $(1)$ we used the fact that $\cos(nx)=\text{Re}(e^{inx})$, then summed the geometric series $\sum_{n=1}^N (-e^{ix})^n$ and took the real part.
In going from $(2)$ to $(3)$ we integrated by parts with $u=\tan(x/2)$ and $v=-\frac{\cos((N+1)x)}{N+1}$.

Taking the limit as $N\to \infty$ in $(3)$ reveals
$$\begin{align}
\sum_{n=1}^\infty \frac{(-1)^n\sin(n)}{n}&=\lim_{N\to \infty}\sum_{n=1}^N \frac{(-1)^n\sin(n)}{n}\\\\
&=-\frac12
\end{align}$$
And we are done!

Tools Used:  Euler's formula, summing a geometric series, and integration by parts.

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\LARGE a)}$

With the
  Abel-Plana Formula:

\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n}\sin\pars{n} \over n} & =
-1 + \sum_{n = 0}^{\infty}\pars{-1}^{n}\,\mrm{sinc}\pars{n} =
-1 + {1 \over 2}\,\mrm{sinc}\pars{0} = -1 + {1 \over 2} = \bbx{-\,{1 \over 2}}
\end{align}

$\ds{\LARGE b)}$
\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n}\sin\pars{n} \over n} & =
\Im\sum_{n = 1}^{\infty}{\pars{-1}^{n}\expo{\ic n} \over n} =
\Im\sum_{n = 1}^{\infty}{\pars{-\expo{\ic}}^{n} \over n} =
-\,\Im\ln\pars{1 + \expo{\ic}}
\\[5mm] & =
-\,\Im\ln\pars{1 + \cos\pars{1} + \sin\pars{1}\,\ic} =
-\arctan\pars{\sin\pars{1} \over 1 + \cos\pars{1}}
\\[5mm] & =
-\arctan\pars{2\sin\pars{1/2}\cos\pars{1/2} \over 2\cos^{2}\pars{1/2}} =
-\arctan\pars{\tan\pars{1 \over 2}} = \bbx{-\,{1 \over 2}}
\end{align}
A: For any $|z|<1$ we have $\sum_{n\geq 1}\frac{(-1)^n z^n}{n}=-\log(1+z)$, hence by picking $z=e^{-u}e^{i}$ with $u>0$ we get
$$ \sum_{n\geq 1}\frac{(-1)^n e^{-nu}\sin(n)}{n} = -\text{Im}\log(1+e^{-u}e^{i})=-\text{Arg}(e^u+e^{i}). $$
The sequences $\{\sin n\}_{n\geq 1}$ and $\{(-1)^n \sin n\}_{n\geq 1}$ have bounded partial sums, hence by summation by parts we are allowed to state that
$$\begin{eqnarray*} \lim_{u\to 0^+}\sum_{n\geq 1}\frac{(-1)^n e^{-nu}\sin(n)}{n} &=&\sum_{n\geq 1}\frac{(-1)^n\sin(n)}{n}\\ &=& -\text{Arg}(1+e^i) = -\frac{1}{2}-\text{Arg}\left(2\cos\tfrac{1}{2}\right)=\color{red}{-\frac{1}{2}}.\end{eqnarray*}$$
