Calc the sum of $\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$ Solving a bigger problem about Fourier series I'm faced with this sum:
$$\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$$
and I've no idea of how to approach this.
I've used Leibniz convergence criterium to verify that the sum should have a value, but I don't know how to calculate this value.
 A: Hint:
$e^{2ik}-e^{-2ik}=2i\sin 2k$
A: $\newcommand{\+}{^{\dagger}}
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Lets $\ds{\fermi\pars{x}\equiv
          \left\{\begin{array}{lcrcl}
                 {\sin\pars{2x} \over x} & \mbox{if} & x & \not= & 0
                 \\[1mm]
                 2 & \mbox{if} & x & = & 0
                 \end{array}\right.}$

Note that $\ds{\fermi}$ is an even function of $\ds{x}$:
  $\ds{\fermi\pars{-x}=\fermi\pars{x}\,,\ \forall\ x\in{\mathbb R}}$.

We'll use the
$\large\mbox{Abel-Plana Formula}$:

\begin{align}
&\color{#66f}{\large\sum_{n = 1}^{\infty}\pars{-1}^{n}\,{\sin\pars{2n} \over n}}
=-2 + \sum_{n = 0}^{\infty}\pars{-1}^{n}\fermi\pars{n}
\\[5mm]&=-2 + \bracks{\half\,\fermi\pars{0} + \ic\
\underbrace{\int_{0}^{\infty}%
{\fermi\pars{\ic t} - \fermi\pars{-\ic t} \over 2\sinh\pars{\pi t}}\,\dd t}
_{\ds{=\ \color{#c00000}{\large 0}}}}
\\[5mm]&=-2 + \bracks{\half\times 2 + \ic\times 0} = \color{#66f}{\Large -1}
\end{align}

A: $$\sum_{k=1}^{+\infty}\frac{(-1)^k}{k}\sin(2k)=\Im\sum_{k=1}^{+\infty}\frac{(-1)^k}{k}e^{2ik}=-\Im\log(1+e^{2i})=\color{red}{-1}.$$
A: Consider the series
\begin{align}
S = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n).
\end{align}
Method 1

Using the known Fourier series
\begin{align}
x = \frac{2 L}{\pi} \, \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \, \sin\left( \frac{n \pi x}{L} \right)
\end{align}
it can quickly be seen that for $L = \pi$ and $x = 2$ the series becomes
\begin{align}
- 1 = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n).
\end{align}
Method 2

Using $2i \sin(2n) = e^{2in} - e^{-2in}$ then the series is
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n) &= \frac{1}{2i} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \left( e^{2in} - e^{-2in} \right) \\
&= - \frac{1}{2i} \left( \ln(1 + e^{2i}) - \ln(1 + e^{-2i}) \right) \\
&= - \frac{1}{2i} \, \ln\left(\frac{1 + e^{2i}}{1 + e^{-2i}} \right) 
= - \frac{1}{2i} \, \ln\left(\frac{e^{i} \, \cos(1)}{e^{-i} \, \cos(1)} \right) \\
&= - \frac{1}{2i} \ln(e^{2i}) = -1.
\end{align}
Method 3

As stated in the proposed problem the summation is given by
\begin{align}
S_{0} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n),
\end{align}
for which 
\begin{align}
S_{0} &= \lim_{n \rightarrow 0} \left\{ \frac{\sin(2n)}{n} \right\} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n) \\
&= -1 + \lim_{n \rightarrow 0} \left\{ \frac{2 \cos(2n)}{1} \right\} \\
&= -1 + 2 = 1.  
\end{align}
