Infinite sum

$\sum_{n=2}^{\infty}\frac1{8n^3-2n}=(\frac{1}{2})\sum_{n=2}^{\infty}[\frac{1}{n(2n-1)(2n+1)}]=(\frac{1}{4})\sum_{n=2}^{\infty}[\frac{1}{n(2n-1)}-\frac{1}{n(2n+1)}]=(\frac{1}{4})\sum_{n=2}^{\infty}[\frac{2}{(2n-1)}-\frac{1}{n}]-[\frac{1}{n}-\frac{2}{(2n+1)}]=(\frac{1}{4})\sum_{n=2}^{\infty}[\frac{2}{(2n-1)}-\frac{2}{n}+\frac{2}{(2n+1)}]=(\frac{1}{2})\sum_{n=2}^{\infty}[\frac{1}{(2n-1)}-\frac{1}{n}+\frac{1}{(2n+1)}]$ $(\frac{1}{2})[\sum_{n=2}^{\infty}\frac{1}{(2n-1)}-\sum_{n=2}^{\infty}\frac{1}{n}+\sum_{n=2}^{\infty}\frac{1}{(2n+1)}]$ $(\frac{1}{2})[\sum_{n=1}^{\infty}\frac{1}{(2n+1)}-\sum_{n=2}^{\infty}\frac{1}{n}+\sum_{n=2}^{\infty}\frac{1}{(2n+1)}]$ $(\frac{1}{2})[\frac{1}{3}+2\sum_{n=2}^{\infty}\frac{1}{(2n+1)}-\sum_{n=2}^{\infty}\frac{1}{n}]$ $(\frac{1}{2})[\frac{1}{3}+2(\sum_{n=2}^{\infty}\frac{1}{(2n+1)}-\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n})]$ $\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...=\sum_{n=1}^{\infty}\frac{({-1})^{n-1}}{n}=log_{e} 2}}}}$ $(\frac{1}{2})[\frac{1}{3}+2(\sum_{n=2}^{\infty}\frac{1}{(2n+1)}-\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n})=(\frac{1}{2})[\frac{1}{3}+2(log_{e} 2-\frac{5}{6})]$ $(\frac{1}{2})[\frac{1}{3}+(2log_{e} 2-\frac{5}{3})]$ $(\frac{1}{2})[(2log_{e} 2-\frac{4}{3})]$ $\rightarrow\Large\color{#3D99F6}{\boxed{(log_{e} 2-\frac{2}{3})}}$

It should be

$\displaystyle -\frac{2}{3} + ln(2)$