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Is there a closed form for the following?$$\sum_{k=-\infty}^{\infty} \frac{(-1)^k}{x+(2k+1)}$$

I know that the non-sign-alternating sum for all integer k (i.e. $\sum \frac{1}{x+k}$)is $\pi cot(\pi x)$. But I am not able to use that result to get what I want.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{k = -\infty}^{\infty}{\pars{-1}^{k} \over x + \pars{2k + 1}} = {1 \over x + 1} + \sum_{k = 1}^{\infty}\bracks{% {\pars{-1}^{-k} \over x + \pars{-2k + 1}} + {\pars{-1}^{k} \over x + \pars{2k + 1}}} \\[5mm] = &\ -\,{1 \over x + 1} + {1 \over 2}\sum_{k = 0}^{\infty}\bracks{% {\pars{-1}^{k} \over k + \pars{x + 1}/2} - {\pars{-1}^{k} \over k - \pars{x + 1}/2}} \\[5mm] = &\ -\,{1 \over 4\xi} + \bracks{{1 \over 2}\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 2\xi} - \pars{~\xi \leftrightarrow -\xi~}}\qquad \mbox{where}\quad \xi \equiv {1 \over 4}\pars{x + 1}.\label{1}\tag{1} \end{align}


Then, \begin{align} {1 \over 2}\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 2\xi} & = {1 \over 2}\sum_{k = 0}^{\infty} \pars{{1 \over 2k + 2\xi} - {1 \over 2k + 1 + 2\xi}} = {1 \over 4}\sum_{k = 0}^{\infty} \pars{{1 \over k + \xi} - {1 \over k + \xi + 1/2}} \\[5mm] & = {H_{\xi - 1/2} - H_{\xi - 1} \over 4}\qquad\pars{~H_{z}:\ Harmonic Number~} \label{1.a}\tag{1.a} \end{align}
\eqref{1} becomes: \begin{align} &\sum_{k = -\infty}^{\infty}{\pars{-1}^{k} \over x + \pars{2k + 1}} = -\,{1 \over 4\xi} + \pars{% {H_{\xi - 1/2} - H_{\xi - 1} \over 4} - {H_{-\xi - 1/2} - H_{-\xi - 1} \over 4}} \\[5mm] = &\ {H_{\xi - 1/2} - H_{-\xi - 1/2} \over 4} - {\pars{H_{\xi - 1} + 1/\xi} - H_{-\xi - 1} \over 4} = {H_{\xi - 1/2} - H_{-\xi - 1/2} \over 4} - {H_{\xi} - H_{-\xi - 1} \over 4} \label{2}\tag{2} \end{align} where I used the $\ds{H_{z}}$ Recursive Property.
With Euler Reflection Formula: $$ \left\{\begin{array}{l} \ds{H_{\xi - 1/2} - H_{-\xi - 1/2} = \pi\cot\pars{\pi\,\bracks{-\xi + {1 \over 2}}} = \bbx{\pi\tan\pars{\pi\xi}}} \\[5mm] \ds{H_{\xi} - H_{-\xi - 1} = \pi\cot\pars{\pi\bracks{-\xi}} = \bbx{-\pi\cot\pars{\pi\xi}}} \end{array}\right. $$
\eqref{2} becomes: \begin{align} \sum_{k = -\infty}^{\infty}{\pars{-1}^{k} \over x + \pars{2k + 1}} & = {1 \over 4}\,\pi\bracks{\tan\pars{\pi\xi} + \cot\pars{\pi\xi}} = {1 \over 4}\,\pi\,{\sec^{2}\pars{\pi\xi} \over \tan\pars{\pi\xi}} = {1 \over 2}\,\pi\,{1 \over \sin\pars{2\pi\xi}} \\[5mm] & = {1 \over 2}\,\pi\, {1 \over \sin\pars{\pi x/2 + \pi/2}} = \bbox[#ffe,15px,border:1px dotted navy]{% {1 \over 2}\,\pi\sec\pars{\pi x \over 2}} \end{align}

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  • $\begingroup$ Lots of new stuff for me here. Thanks for this answer. I might need more clarification on some steps. Firstly, is $\sum_{0}^{\infty} \frac{1}{k+x}$ equal to $H_{x-\frac{1}{2}}$? $\endgroup$ – Srini Jul 3 '17 at 5:23
  • $\begingroup$ @Srini $\color{#f00}{NO}$. $\sum_{k = 0}^{n}{1 \over k + x} = H_{x + n} -H_{x - 1}$ which diverges as $n \to \infty$. You can see, in $\left(1.a\right)$, that the difference of two 'similar' ones converge. Thanks for your remark. $\endgroup$ – Felix Marin Jul 3 '17 at 21:10
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Split it up:

$$\sum_{k=-\infty}^\infty\frac{(-1)^k}{x+2k+1}=\sum_{k=-\infty}^\infty\underbrace{\frac1{x+4k+1}}_{S_1}-\underbrace{\frac1{x+4k+3}}_{S_2}$$

$$S_1=\sum_{k=-\infty}^\infty\frac1{4k+x+1}=\frac14\sum_{k=-\infty}^\infty\frac1{k+\frac{x+1}4}=\frac\pi4\cot\left(\frac{x+1}4\pi\right)$$

$$S_2=\sum_{k=-\infty}^\infty\frac1{4k+x+3}=\frac14\sum_{k=-\infty}^\infty\frac1{k+\frac{x+3}4}=\frac\pi4\cot\left(\frac{x+3}4\pi\right)$$

$$\sum_{k=-\infty}^\infty\frac{(-1)^k}{x+2k+1}=\frac\pi4\left[\cot\left(\frac{x+1}4\pi\right)+\tan\left(\frac{x+1}4\pi\right)\right]=\frac\pi2\sec\left(\frac{\pi x}2\right)$$

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    $\begingroup$ @FelixMarin Whoops, fixed it. $\endgroup$ – Simply Beautiful Art Jul 1 '17 at 23:59
  • $\begingroup$ @FelixMarin Nah, its fine, you're supposed to bother me when I make these silly mistakes lol $\endgroup$ – Simply Beautiful Art Jul 2 '17 at 11:38
  • $\begingroup$ One should justify rearranging conditionally convergent series. $\endgroup$ – GEdgar Jul 2 '17 at 12:07
  • $\begingroup$ @GEdgar It's easy to see from the partial sums that no rearrangement here has changed the end result. $\endgroup$ – Simply Beautiful Art Jul 2 '17 at 12:09
  • $\begingroup$ Thank you for mentioning it. $\endgroup$ – GEdgar Jul 2 '17 at 12:13
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From the residue theorem we know that $$\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}f\left(k\right)=-\sum\textrm{residues of }\pi\csc\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{ poles}$$ where $f\left(z\right)$ verifies some hypotheses (see here). So since we have a simple pole at $z=\left(-x-1\right)/2,$ we get $$\sum_{k\in\mathbb{Z}}\frac{\left(-1\right)^{k}}{x+2k+1}=-\underset{z=\left(-x-1\right)/2}{\textrm{Res}}\frac{\pi\csc\left(\pi z\right)}{2z+1+x}=\color{red}{\frac{\pi}{2}\sec\left(\frac{\pi}{2}x\right)}$$ as wanted.

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