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How to compute the following series:

$$\sum_{n=1}^{\infty}\frac{1}{16n^2-1}$$

I tried to use partial fraction

$$\frac{1}{16n^2-1}=\frac{1}{(4n-1)(4n+1)}=\frac{1}{2}\left[\frac{1}{4n-1}-\frac{1}{4n+1}\right]$$

I thought it will form telescoping series, but it is not. Any help would be appreciated. Thanks in advance.

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  • $\begingroup$ If you write down the terms, $$\frac{1}{3} - \frac{1}{5} + \frac{1}{7} - \frac{1}{9} + \frac{1}{11} - \dotsc,$$ you may be reminded of the Leibniz series. $\endgroup$ – Daniel Fischer May 26 '14 at 21:20
  • $\begingroup$ @DanielFischer Thanks for your suggestion. :) $\endgroup$ – Venus May 26 '14 at 21:28
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Rewrite \begin{align} \sum_{n=1}^\infty\frac1{16n^2-1}&=\sum_{n=1}^\infty\frac1{(4n-1)(4n+1)}\\ &=\sum_{n=1}^\infty\frac12\left(\frac1{4n-1}-\frac1{4n+1}\right)\\ &=\frac12\left(\frac{1}{3}-\frac{1}5+\frac{1}7-\frac{1}9+\frac{1}{11}-\frac1{13}+\cdots\right). \end{align} Recall Leibniz series $$ \frac\pi4=1-\frac13+\frac15-\frac17+\frac19-\frac1{11}+\cdots $$ then $$ \sum_{n=1}^\infty\frac1{16n^2-1}=\color{blue}{\frac12\left(1-\frac\pi4\right)}. $$

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    $\begingroup$ Thank you for your answer. :) $\endgroup$ – Venus May 26 '14 at 21:29
  • $\begingroup$ @Tunk-Fey, how did you get from $\frac{1}{(4n-1)(4n+1)}$ to $1/2(\frac{1}{4n-1}-\frac{1}{4n+1})$? $\endgroup$ – recursive recursion May 26 '14 at 21:48
  • $\begingroup$ @recursiverecursion It's partial fraction. See the OP. $\endgroup$ – Tunk-Fey May 26 '14 at 21:52
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    $\begingroup$ @recursiverecursion $$ \frac{1}{(4n-1)(4n+1)}=\frac12\cdot\frac{2}{(4n-1)(4n+1)}=\frac12\cdot\frac{(4n+1)-(4n-1)}{(4n-1)(4n+1)} $$ $\endgroup$ – Tunk-Fey May 26 '14 at 21:54
  • $\begingroup$ Oh, alright. Sorry, didn't see that immediately. $\endgroup$ – recursive recursion May 26 '14 at 22:02
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Generally speaking, $~\displaystyle\sum_{n=-\infty}^\infty\dfrac{x^2}{x^2-n^2}=\pi x\cdot\cot(\pi x)$. In this case, $x=\dfrac14$ . This can be proven by

taking the natural logarithm of Euler's infinite product expression for $\dfrac{\sin(\pi x)}{\pi x}$, and differentiating

both sides. If the minus sign in the denominator would be changed to a plus sign, then the series

would become $\pi x\cdot\coth(\pi x)$.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{1 \over 16n^{2} - 1}:\ {\large ?}}$

Even though the result is quite trivial, I was curious about the factor $\ds{\pi \over 4}$ which is equal to $\ds{\arctan\pars{1} = \int_{0}^{1}{\dd x \over x^{2} + 1}}$. So, this integral should be involved in the calculation. That's what I want to show here which adds another way to arrive to the result.

\begin{align} \color{#66f}{\large\sum_{n = 1}^{\infty}{1 \over 16n^{2} - 1}}& =\half\pars{{1 \over 4n - 1} - {1 \over 4n + 1}} =-\,{1 \over 8}\sum_{n = 1}^{\infty}\sum_{\sigma = \pm}{\sigma \over n + \sigma/4} \\[3mm]&=-\,{1 \over 8}\sum_{n = 1}^{\infty}\sum_{\sigma = \pm}\sigma \int_{0}^{1}t^{n + \sigma/4 - 1}\,\dd t =-\,{1 \over 8}\sum_{\sigma = \pm}\sigma\int_{0}^{1} \sum_{n = 1}^{\infty}t^{n + \sigma/4 - 1}\,\dd t \\[3mm]&=-\,{1 \over 8}\sum_{\sigma = \pm}\sigma \int_{0}^{1}{t^{\sigma/4} \over 1 - t}\,\dd t ={1 \over 8}\int_{0}^{1}{1 - t^{1/2} \over t^{1/4}\pars{1 - t}}\,\dd t \\[3mm]&={1 \over 8}\ \overbrace{\int_{0}^{1}{\dd t \over t^{1/4}\pars{1 + t^{1/2}}}} ^{\ds{\mbox{Set}\ t = x^{4}}}\ =\ \half\int_{0}^{1}{x^{2} \over 1 + x^{2}}\,\dd x \\[3mm]&=\half\pars{\int_{0}^{1}\dd x - \int_{0}^{1}{\dd x \over 1 + x^{2}}} =\half\bracks{1 - \arctan\pars{1}} =\color{#66f}{\large\half\pars{1 - {\pi \over 4}}} \end{align}

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    $\begingroup$ @Venus You're welcome. Thanks. $\endgroup$ – Felix Marin Jun 11 '14 at 18:03

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