Question

Найдите пожалуйста сумму ряда

Answer 1

$\displaystyle \sum^\infty_{n=1}\dfrac{1}{n(n+5)}=\dfrac{1}{5}\sum^\infty_{n=1}\dfrac{n+5-n}{n(n+5)}=\dfrac{1}{5}\left(\sum^\infty_{n=1}\dfrac{1}{n}-\sum^\infty_{n=1}\dfrac{1}{n+5}\right)=$

$\displaystyle \dfrac{1}{5}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\sum^\infty_{n=6}\dfrac{1}{n}-\sum^\infty_{n=1}\dfrac{1}{n+5}\right)=\dfrac{1}{5}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\right)=\dfrac{137}{300}$