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.