Sum $\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)}$ Find the value of
$$\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)}$$
I have tried to substitute the value of $n$ from $1$ but it is not telescoping series.

Hint. Recall the following series representation for the digamma function
$$\begin{equation} 
\psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}  
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant. 
Then by partial fraction decomposition we have
$$
\begin{align}
\frac{1}{(9n-1)(9n+1)} &= \frac{1}{2}\left(\frac{1}{9n-1}-\frac{1}{9n+1}\right)\\\\
&=\frac{1}{18}\left(\frac{1}{n-1/9}-\frac{1}{n+1/9}\right)\\\\
&=\frac{1}{18}\left[\left(\frac{1}{n-1/9}-\frac1n\right)-\left(\frac{1}{n+1/9}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} =\frac{1}{18}\left(\psi\left(\frac{10}{9}\right)-\psi\left(\frac{8}{9}\right)\right)
$$ equivalently

$$
\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} =\frac{1}{2}-\frac{\pi}{18}\cot\left(\frac{\pi}{9}\right)
$$ 

where we have used Gauss' theorem for the digamma function. A numerical approximation for this series is given by $$\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} \approx \color{blue}{0.02047472906319532...}$$
