Kris Patton

2022-02-24

$\sum _{n=2}^{\mathrm{\infty }}\frac{2}{\left({n}^{3}-n\right){3}^{n}}=-\frac{1}{2}+\frac{4}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\cdot {3}^{n}}$

junoon363km

$\sum _{n=2}^{\mathrm{\infty }}\frac{2}{{3}^{n}\left({n}^{3}-n\right)}=\sum _{n=2}^{\mathrm{\infty }}\frac{1}{{3}^{n}}\left(\frac{-2}{n}+\frac{1}{n+1}+\frac{1}{n-1}\right)$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{3}^{n+1}n}-\sum _{n=1}^{\mathrm{\infty }}\frac{2}{{3}^{n+1}\left(n+1\right)}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+2\right){3}^{n+1}}$
$=\frac{1}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{3}^{n}}-2\left(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{3}^{n}}-\frac{1}{3}\right)+3\left(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{3}^{n}}-\frac{1}{3}-\frac{1}{{3}^{2}\cdot 2}\right)$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{3}^{n}}\left(\frac{1}{3}-2+3\right)+\left(\frac{2}{3}-1-\frac{1}{6}\right)$
$=\frac{4}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\cdot {3}^{n}}-\frac{1}{2}$

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