On the series $\sum \limits_{n=1}^{\infty} \frac{1}{n^2-3n+3}$ and $\sum\limits_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3}$ 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. 
 A: $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
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.
A: 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 $(5)$-$(8)$
