How to find $\sum\limits_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac{1}{2n}}$? How to find $$\sum\limits_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac{1}{2n}}$$
I don't even know where to start. It looks like something more advanced than my current level. Could you please give me at least some starting points? Thank you!
 A: $$\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{k+\frac{1}{2n}}=2n\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{2nk+1}=2n\bigg(\sum_{k=-\infty,k\neq0}^{\infty}(-1)^k\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{(2nk)^j}\bigg)+2n=\\2n\bigg(\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{(2n)^j}\sum_{k=-\infty,k\neq0}^{\infty}\frac{(-1)^k}{k^j}\bigg)+2n$$
The inner sum is $0$ if $j$ is odd and $(2^{2-j}-2)\zeta(j)$ otherwise. ($\zeta(j)$ represents the Riemann Zeta function, of course) $$2n\sum_{j=1}^{\infty}\frac{(-1)^{2j+1}(2^{2-2j}-2)\zeta(2j)}{(2n)^{2j}}+2n=-2n\sum_{j=1}^{\infty}\frac{(2^{2-2j}-2)\zeta(2j)}{(2n)^{2j}}+2n$$ 
Luckily, the even values of the Zeta function have a closed form- specifically,
$\zeta(2j)=\frac{(-1)^{j+1}B_{2j}(2\pi)^{2j}}{2(2j)!}$  as can be found here.    
So the sum can be rewritten as:  
$$-2n\sum_{j=1}^{\infty}\frac{(-1)^{j+1}B_{2j}(2\pi)^{2j}(2^{2-2j}-2)}{2(2j)!(2n)^{2j}}+2n\\=-4n\sum_{j=0}^{\infty}\frac{(-1)^jB_{2j}(2\pi)^{2j}}{2(2j)!(2n)^{2j}}+8n\sum_{j=0}^{\infty}\frac{(-1)^jB_{2j}(2\pi)^{2j}}{2(2j)!(2n)^{2j}2^{2j}}\\=-2n\sum_{j=0}^{\infty}\frac{(-1)^jB_{2j}2^{2j}(\frac{\pi}{2n})^{2j}}{(2j)!}+4n\sum_{j=0}^{\infty}\frac{(-1)^jB_{2j}2^{2j}(\frac{\pi}{4n})^{2j}}{(2j)!}\\=-\pi\sum_{j=0}^{\infty}\frac{(-1)^jB_{2j}2^{2j}(\frac{\pi}{2n})^{2j-1}}{(2j)!}+\pi\sum_{j=0}^{\infty}\frac{(-1)^jB_{2j}2^{2j}(\frac{\pi}{4n})^{2j-1}}{(2j)!}\\=-\pi\operatorname{cot}\bigg(\frac{\pi}{2n}\bigg)+\pi\operatorname{cot}\bigg(\frac{\pi}{4n}\bigg)\\=\frac{\pi}{\operatorname{sin}(\frac{\pi}{2n})}$$ 
(that last series used is the Taylor series expansion for cotangent which you can also find here)  
(please edit or comment for any corrections)
A: Let $0<x<1$. Define $f_x$ be the $2\pi$-periodic function defined by $f_x(t)=\cos(xt)$ for $|t|< \pi$.
You can verify that its Fourier series expansion is
$$\cos{xt} = \frac{\sin{\pi x}}{\pi x} \left [1+2 x^2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \cos{k t}}{k^2-x^2} \right ]$$
Evaluating at $t=0$ yields
$$\begin{split}1 &= \frac{\sin{\pi x}}{\pi x} \left [1+2 x^2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1} }{k^2-x^2} \right ]\\
&=\frac{\sin{\pi x}}{\pi x} \left [1+x \sum_{k=1}^{\infty} (-1)^{k+1}\left(\frac{1 }{k-x} -\frac 1 {k+x}\right)\right ]\\
&= \frac{\sin{\pi x}}{\pi x} \left [x \sum_{k\in\mathbb Z} \frac{(-1)^{k}}{k+x} \right]\\
\end{split}$$
Conclusion: For all $x\in(0, 1)$,
$$\sum_{k\in\mathbb Z} \frac{(-1)^{k}}{k+x} = \frac{\pi}{\sin(\pi x)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{k = -\infty}^{\infty}{\pars{-1}^{k} \over k + 1/\pars{2n}}} =
2n + \sum_{k = 1}^{\infty}\bracks{{\pars{-1}^{k} \over k + 1/\pars{2n}} +
{\pars{-1}^{-k} \over -k + 1/\pars{2n}}}
\\[5mm] = &\
2n - \sum_{k = 0}^{\infty}\bracks{%
{\pars{-1}^{k} \over k + 1 + 1/\pars{2n}} -
{\pars{-1}^{k} \over k + 1 - 1/\pars{2n}}}
\\[1cm] = &\
2n - \sum_{k = 0}^{\infty}\left\{%
\bracks{{1 \over 2k + 1 + 1/\pars{2n}} -
{1 \over 2k + 1 - 1/\pars{2n}}}\right.
\\[2mm] & \phantom{2n - \sum_{k = 0}^{\infty}\,}-
\left.\bracks{{1 \over 2k + 2 + 1/\pars{2n}} -
{1 \over 2k + 2 - 1/\pars{2n}}}\right\}
\\[1cm] = &\
2n - {1 \over 2}\sum_{k = 0}^{\infty}
\bracks{{1 \over k + 1/2 + 1/\pars{4n}} -
{1 \over k + 1/2 - 1/\pars{4n}}}
\\[2mm] & \phantom{2n\,}
+ {1 \over 2}\sum_{k = 0}^{\infty}\bracks{{1 \over k + 1 + 1/\pars{4n}} -
{1 \over k + 1 - 1/\pars{4n}}}
\\[1cm] = &\
2n - {1 \over 2}\bracks{\Psi\pars{{1 \over 2} - {1 \over 4n}} -
\Psi\pars{{1 \over 2} + {1 \over 4n}}} +
{1 \over 2}\bracks{\Psi\pars{1 - {1 \over 4n}} -
\Psi\pars{1 + {1 \over 4n}}}
\\[1cm] = &\
2n - {1 \over 2}\pi\cot\pars{\pi\bracks{{1 \over 2} + {1 \over 4n}}}
+
\braces{{1 \over 2}\bracks{\Psi\pars{1 - {1 \over 4n}} -
\Psi\pars{1 \over 4n}} - {1 \over 2}\,{1 \over 1/\pars{4n}}}
\\[5mm] = &\
{1 \over 2}\,\pi\tan\pars{\pi \over 4n} +
{1 \over 2}\,\pi\cot\pars{\pi \over 4n} =
\bbx{\pi\csc\pars{\pi \over 2n}}
\end{align}
