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So, in order to obtain the required answer, I tried to apply some Taylor expansions, which led me to nowhere actually. After a while I tried to use the summation theorem

$\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}}$ at $f\left(z\right)$'s poles.

Residue at $z=-\frac{1}{2}$ equals to residue at $z=\frac{1}{2}$, both of them are $-\frac{\pi ^2}{16}$. What I got seems to be not the correct answer after all:

$-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}} = -\frac{\pi ^2}{8}$, so that $\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{8}$ Since the $f\left(z\right)$ is even, I get $\sum_{n=0}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{16}$

However, the initial task was to find the sum of $\sum_{n=1}^{+\infty}{f\left(n\right)}$. I supposed that $\sum_{n=1}^{+\infty}{f\left(n\right)}=\sum_{n=0}^{+\infty}{f\left(n\right)}-f\left(0\right)=\frac{\pi ^2}{16}-1$, which is incorrect, as according to Mathematica, I should get $\frac{\pi ^2}{16}-\frac{1}{2}$. I don't know where is my mistake at this point. Perhaps, the whole approach is not well executed.

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2 Answers 2

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Since $\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)$ by squaring we get $$\sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n+1)^2}-\frac{1}{4}\sum_{n\geq 1}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$ where the first two series in the RHS depend on $\zeta(2)-\frac{1}{4}\zeta(2)=\frac{\pi^2}{8}$ and the last one is telescopic: $$ \sum_{n\geq 1}\frac{1}{(4n^2-1)^2}=\color{blue}{\frac{\pi^2-8}{16}}.$$

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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_{n = 1}^{\infty}{1 \over \pars{4n^{2} - 1}^{2}} & = \left.{1 \over 16}\,\partiald{}{a}\sum_{n = 1}^{\infty}{1 \over n^{2} - a^{2}} \right\vert_{\ a\ =\ 1/2} = \left.{1 \over 16}\,\partiald{}{a}\sum_{n = 0}^{\infty}\pars{% {1 \over n + 1 - a} - {1 \over n + 1 + a}}{1 \over 2a} \right\vert_{\ a\ =\ 1/2} \\[5mm] & = \left.{1 \over 32}\,\partiald{}{a}\pars{H_{a} - H_{-a} \over a} \right\vert_{\ a\ =\ 1/2}\qquad\qquad\quad\pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] & = \left.{1 \over 32}\,\partiald{}{a}\pars{H_{a - 1} + 1/a - H_{-a} \over a} \right\vert_{\ a\ =\ 1/2}\qquad\pars{~H_{z}\ Recursive\ Property~} \\[5mm] & = \left.{1 \over 32}\,\partiald{}{a}\pars{{1 \over a^{2}} - {H_{-a} - H_{a - 1} \over a}}\right\vert_{\ a\ =\ 1/2} \\[5mm] & = \left.{1 \over 32}\,\partiald{}{a}\pars{{1 \over a^{2}} - {\pi\cot\pars{\pi a} \over a}}\right\vert_{\ a\ =\ 1/2} \qquad\pars{~H_{z}\ Euler\ Reflection\ Formula~} \\[5mm] & = \left.{1 \over 32}\pars{-\,{2 \over a^{3}} + {\pi\cot\pars{\pi a} \over a^{2}} + {\pi^{2}\csc^{2}\pars{\pi a} \over a}}\right\vert_{\ a\ =\ 1/2} = \bbx{\pi^{2} - 8 \over 16} \approx 0.1169 \end{align}

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