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Wolfram Alpha says that

$$\sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} = 1 + \frac{\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$

However I am unable to get it. It is fairly routine to prove that

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} = \frac{2\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$

by using complex analysis ( contour integration ) but honestly I am stuck how to retrieve the original sum. Split up , the last sum gives:

\begin{align*} \sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} &= \sum_{n=-\infty}^{-1} \frac{1}{n^2-3n+3} + \frac{1}{3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\ &=\frac{1}{3} +\sum_{n=1}^{\infty} \frac{1}{n^2+3n+3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\ &=\frac{1}{3}+ \sum_{n=1}^{\infty} \left [ \frac{1}{n^2-3n+3} + \frac{1}{n^2+3n+3} \right ] \end{align*}

Am I overlooking something here?

P.S: Working with digamma on the other hand I am not getting the constant. I'm getting $\frac{1}{3}$ instead.

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    $\begingroup$ $$\sum_{n=1}^{\infty} \frac{1}{n^2+3n+3}=\sum_{n=1}^{\infty} \frac{1}{n(n+3)+3}=\sum_{k=4}^{\infty} \frac{1}{(k-3)k+3}=\sum_{n=1}^{\infty} \frac{1}{n^2-3n+3}-1-1-\frac13$$ $\endgroup$ – Did Mar 11 '18 at 20:42
  • $\begingroup$ For sure. Thanks @Did $\endgroup$ – Tolaso Mar 11 '18 at 20:46
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    $\begingroup$ If $f(n)=n^2+3n+3,$ then $f(1-n)=f(n)$ $\endgroup$ – saulspatz Mar 11 '18 at 20:57
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$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$ for any $a\neq b$ in the half-plane $\text{Re}(s)>0$ is fairly routine, too. Here we have to find $$ \sum_{n\geq 0}\frac{1}{n^2-n+1}=1+\sum_{n\geq 0}\frac{1}{n^2+n+1}=1+\frac{\psi\left(\frac{1+i\sqrt{3}}{2}\right)-\psi\left(\frac{1-i\sqrt{3}}{2}\right)}{i\sqrt{3}}\tag{2}$$ which (by the reflection formula for the $\psi$ function) simplifies into $$ 1+\frac{-\pi\cot\left(\frac{\pi}{2}(1+i\sqrt{3})\right)}{i\sqrt{3}}=1+\frac{\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)\approx 2.79814728\tag{3}$$ as wanted.

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The First Sum $$ \begin{align} \sum_{n=1}^\infty\frac1{n^2-3n+3} &=1+\sum_{n=2}^\infty\frac1{n^2-3n+3}\tag1\\ &=1+\sum_{n=2}^\infty\frac1{\left(n-\frac32-i\frac{\sqrt3}2\right)\left(n-\frac32+i\frac{\sqrt3}2\right)}\tag2\\ &=1+\frac1{i\sqrt3}\sum_{n=2}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}-\frac1{n-\frac32+i\frac{\sqrt3}2}\right)\tag3\\ &=1+\frac1{i\sqrt3}\sum_{n=2}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}+\frac1{-n+\frac32-i\frac{\sqrt3}2}\right)\tag4\\ &=1+\frac1{i\sqrt3}\sum_{n=-\infty}^\infty\frac1{n+\frac12-i\frac{\sqrt3}2}\tag5\\ &=1+\frac\pi{i\sqrt3}\cot\left(\frac\pi2-i\frac{\pi\sqrt3}2\right)\tag6\\[3pt] &=1+\frac\pi{i\sqrt3}\tan\left(i\frac{\pi\sqrt3}2\right)\tag7\\[3pt] &=1+\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)\tag8 \end{align} $$ Explanation:
$(1)$: separate the $n=1$ term
$(2)$: factor the denominator
$(3)$: apply partial fractions
$(4)$: rewrite the right hand summand
$(5)$: combine the summands into a sum over $\mathbb{Z}$
$(6)$: apply $(7)$ from this answer
$(7)$: $\cot\left(\frac\pi2-x\right)=\tan(x)$
$(8)$: $\tan(ix)=i\tanh(x)$


The Second Sum $$ \begin{align} \sum_{n=-\infty}^\infty\frac1{n^2-3n+3} &=\frac1{i\sqrt3}\sum_{n=-\infty}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}+\frac1{-n+\frac32-i\frac{\sqrt3}2}\right)\tag9\\ &=\frac2{i\sqrt3}\sum_{n=-\infty}^\infty\frac1{n+\frac12-i\frac{\sqrt3}2}\tag{10}\\ &=\frac2{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)\tag{11} \end{align} $$ Explanation:
$\phantom{1}(9)$: partial fractions á la $(3)$
$(10)$: combine two series over $\mathbb{Z}$
$(11)$: apply $(5)$-$(8)$

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  • $\begingroup$ Thanks Robjohn ! $\endgroup$ – Tolaso Mar 11 '18 at 22:20

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