# 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!

• Haven't worked out details - but my first idea would be to apply complex analysis to relate residues of $f(z) = \frac{\csc(\pi z)}{z + 1/(2n)}$ and if that works, then my guess would be the answer is something along the lines of $\csc(\frac{\pi}{2n})$. – Daniel Schepler Feb 19 at 19:51
• You have 2 answers with, apparently, different results. – DanielWainfleet Feb 20 at 6:39

Let $$0. Define $$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 $$x\in(0, 1)$$, $$\sum_{k\in\mathbb Z} \frac{(-1)^{k}}{k+x} = \frac{\pi}{\sin(\pi x)}$$

$$\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{k+\frac{1}{2n}}=2n\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{2nk+1}=2n\bigg(\sum_{k=-\infty,k\neq0}^{\infty}(-1)^k\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{(2nk)^j}\bigg)+2n=\\2n\bigg(\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{(2n)^j}\sum_{k=-\infty,k\neq0}^{\infty}\frac{(-1)^k}{k^j}\bigg)+2n$$
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)

• I am not sure concerning the result. Using the Hurwitz-Lerch transcendent function, we arrive to $\pi \csc \left(\frac{\pi }{2 n}\right)$ – Claude Leibovici Feb 20 at 7:47
• @ClaudeLeibovici Yeah I'm not so sure about this either. I may take this down soon. – Cardioid_Ass_22 Feb 20 at 7:52

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