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.
 A: 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 $$\sum_{n=1}^{\infty}\frac{1}{(9n-1)(9n+1)} \approx \color{blue}{0.02047472906319532...}$$
A: In general, for natural values of a and b we have $~\displaystyle\sum_{n=1}^\infty\frac1{(n+a)(n+b)}=\frac{H_a-H_b}{a-b}$ , which 
can easily be proven by decomposing the summand into partial fractions, and noticing that 
the series telescopes. However, in this case, a and b are not integers, but rather $\pm~\dfrac19$ , so the 
series does not telescope. However, that is ultimately of no consequence, since, three centuries 
ago, the great Leonhard Euler has found a way of generalizing harmonic numbers to non–
natural arguments, so the same formula applies ! Of course, the integrals are messy, requiring 
a substitution of the form $t=\sqrt[\Large^9]x$ , followed by partial fraction decomposition. You will then 
encounter difficulty evaluating $\displaystyle\int_0^1\frac{2~t^4+t^3+t+2}{1+t^3+t^6}dt$ , where you'll have to very graciously 
tip-toe around the three cube roots of each of the two complex cube roots of unity. But, with 
patience, and with the help of complex logarithms and Euler's formula, even this hurdle will 
be overcome.
A: We can exploit the Weierstrass product for the sine function. Since:
$$\frac{\sin(\pi z)}{\pi z} = \prod_{n\geq 1}\left(1-\frac{z^n}{n^2}\right)\tag{1}$$
by considering the logarithmic derivatives of $(1)$ it follows that:
$$ -\frac{1}{z}+\pi\cot(\pi z) = \sum_{n\geq 1}\frac{2z}{z^2-n^2}\tag{2}$$
and we just need to plug in $z=\frac{1}{9}$ to get:
$$\sum_{n\geq 1}\frac{1}{81n^2-1}= \color{red}{\frac{1}{2}-\frac{\pi}{18}\,\cot\frac{\pi}{9}}.\tag{3}$$
$\cot\frac{\pi}{9}$ is an algebraic number of degree six over $\mathbb{Q}$:
$$\cot\frac{\pi}{9}=\sqrt{3+\frac{8}{\sqrt{3}}\,\cos\frac{\pi}{18}}.\tag{4} $$
By expanding the terms of the LHS of $(3)$ as geometric series we also have:
$$\sum_{n\geq 1}\frac{1}{81n^2-1}=\frac{\zeta(2)}{3^4}+\frac{\zeta(4)}{3^8}+\ldots = \sum_{m\geq 1}\frac{\zeta(2m)}{3^{4m}}.\tag{5}$$
