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In an exercise, I caculate the Fourier expansion of $e^x$ over $[0,\pi]$ is $$e^x\sim \frac{e^\pi-1}{\pi}+\frac{2(e^\pi-1)}{\pi}\sum_{n=1}^\infty \frac{\cos 2nx}{4n^2+1}+\frac{4(1-e^\pi)}{\pi}\sum_{n=1}^\infty \frac{n\sin 2nx}{4n^2+1}.$$ From this, it is easy to deduce $$\sum_{n=1}^\infty \frac{1}{4n^2+1}=\frac{\pi}{4}\frac{e^\pi+1}{e^\pi-1}-\frac{1}{2}.$$ However, I could not find the following sum $$\sum_{n=1}^\infty \frac{1}{(2n-1)^2+1},$$ from which we can calculate the sum $\sum 1/(n^2+1)$.

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  • $\begingroup$ use residue theorem ... $\endgroup$ – Math-fun Dec 16 '14 at 8:40
  • $\begingroup$ A related question. $\endgroup$ – Lucian Dec 16 '14 at 8:41
  • $\begingroup$ @Mehdi Just use math analysis... $\endgroup$ – xldd Dec 16 '14 at 9:27
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We can approach such kind of series by considering logarithmic derivatives of Weierstrass products. For instance, from: $$\cosh z = \prod_{n=1}^{+\infty}\left(1+\frac{4z^2}{(2n-1)^2\pi^2}\right)\tag{1}$$ we get: $$\frac{\pi}{2}\tanh\frac{\pi z}{2} = \sum_{n=1}^{+\infty}\frac{2z}{z^2+(2n-1)^2}\tag{2},$$ so, evaluating in $z=1$: $$\sum_{n=1}^{+\infty}\frac{1}{(2n-1)^2+1}=\color{red}{\frac{\pi}{4}\tanh\frac{\pi}{2}}.\tag{3}$$ With the same approach, but starting from the Weierstrass product for $\frac{\sinh z}{z}$, we can compute $\sum_{n\geq 1}\frac{1}{1+n^2}$, too: $$\sum_{n\geq 1}\frac{1}{n^2+1}=\frac{-1+\pi\coth\pi}{2}.\tag{4}$$

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    $\begingroup$ I like it! Excellent! $\endgroup$ – Math-fun Dec 16 '14 at 10:41
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2} + 1}} =\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1 + \ic}\pars{2n - 1 - \ic}} \\[5mm]&=\sum_{n\ =\ 0}^{\infty}{1 \over \pars{2n + 1 + \ic}\pars{2n + 1 - \ic}} ={1 \over 4}\sum_{n\ =\ 0}^{\infty} {1 \over \bracks{n + \pars{1 + \ic}/2}\bracks{n + \pars{1 - \ic}/2}} \\[5mm]&={1 \over 4}\, {\Psi\pars{\bracks{1 + \ic}/2} - \Psi\pars{\bracks{1 - \ic}/2}\over \pars{1 + \ic}/2 - \pars{1 - \ic}/2} \end{align} where $\ds{\Psi}$ is the Digamma Function.

Then, $$ \color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2} + 1}} ={1 \over 4}\,\ic\bracks{\Psi\pars{1 - \ic \over 2} - \Psi\pars{1 + \ic \over 2}} $$

With Euler Reflection Formula $\ds{\Psi\pars{1 - x} - \Psi\pars{x} = \pi\cot\pars{\pi x}}$: \begin{align}&\color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2} + 1}} ={1 \over 4}\,\ic\ \pi\cot\pars{\pi\,{1 + \ic \over 2}} =-\,{1 \over 4}\,\ic\,\pi\tan\pars{\pi\ic \over 2} \\[5mm]&=-\,{1 \over 4}\,\ic\,\pi\bracks{\ic\tanh\pars{\pi \over 2}} =\color{#66f}{\large{1 \over 4}\,\pi\tanh\pars{\pi \over 2}} \end{align}

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