where this series converges Given the series $$\sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}$$
I am completely stuck and do not  understand the answer from my book which is $\pi^2/36-1$. I need explanation and different approach how this result is gained. Thanks  

Rewrite your series as :
\begin{align}
\tag{1}\sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}&=\sum_{j=0}^{\infty}\frac{1}{(3j-1)(2j-1)}\\
&=\sum_{j=0}^{\infty}\frac2{2j-1}-\frac3{3j-1}\\
\tag{2}&=-2+3+\sum_{j=1}^{\infty}\frac1{j-\frac 12}-\frac1{j-\frac 13}\\
&=1-\psi\left(1-\frac 12\right)+\psi\left(1-\frac 13\right)\\
\tag{3}&=1+2\ln(2)-\frac{3\ln(3)}2+\frac{\pi}{2\sqrt{3}}\\
&\approx 1.645275610234835007\\
\end{align}
using the special values of the digamma function (or the Gauss digamma sum) and the resolution method exposed in the excellent Abramowitz and Stegun $(6.8)$.
Here a more 'elementary' derivation is possible if we observe that for any integer $n>1$ :
\begin{align}
\sum_{j=1}^{\infty}\frac1{j-\frac 1n}-\frac1{j}&=\sum_{j=1}^{\infty}\int_0^1x^{j-\frac 1n-1}-x^{j-1}\;dx\\
&=\int_0^1 \sum_{k=0}^{\infty}\left(x^{k-1/n}-x^{k}\right)\;dx\\
&=\int_0^1 \frac{x^{-1/n}-1}{1-x}\;dx\\
\text{setting}\ x:=y^n\ \text{ we get :}\\
&=n\int_0^1 \frac{y^{-1}-1}{1-y^n}y^{n-1}\;dy\\
&=n\int_0^1 \frac{y^{n-2}-y^{n-1}}{1-y^n}\;dy\\
\text{that may be solved using}&\text{ partial fractions.}
\end{align}
From $(2)$ we need :
\begin{align}
\sum_{j=1}^{\infty}\frac1{j-1/2}-\frac1{j-1/3}&=2\int_0^1 \frac{1-y}{1-y^2}\;dy-3\int_0^1 \frac{y-y^2}{1-y^3}\;dy\\
&=2\ln(2)-3\int_0^1 \frac{y}{1+y+y^2}\;dy\\
&=2\ln(2)-\frac 32\left(\int_0^1 \frac{1+2y}{1+y+y^2}\;dy-\int_0^1 \frac 1{(3/4)+(y+1/2)^2}\;dy\right)\\
&=2\ln(2)-\frac 32\left|\ln(1+y+y^2)-\frac 2{\sqrt{3}}\arctan\left(\frac{1+2y}{\sqrt{3}}\right)\right|_0^1\\
&=2\ln(2)-\frac {3\ln(3)}2+\sqrt{3}\left(\arctan\left(\sqrt{3}\right)-\arctan\left(\frac1{\sqrt{3}}\right)\right)\\
\end{align}
Adding $1$ from $(2)$ we get again the result $(3)$ :
$$\boxed{\displaystyle \sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}=1+2\ln(2)-\frac{3\ln(3)}2+\frac{\pi}{2\sqrt{3}}}$$
