Question

i know (6)/(n(n+2)) converges

i tried to find the sum and i got 2 but it says my answer is wrong.

i know the sum, is by p-series because E(summation)= (1)/(n^2)

Answer 1

now for the sum i got 4.4 as the unchanging sum

Answer 2

\[\frac{6}{n(n+2)}=3\left(\frac{1}{n}-\frac{1}{n+2}\right)\]\[\sum\limits_{n=1}^{+\infty}\frac{6}{n(n+2)}=3\left(\sum\limits_{n=1}^{+\infty}\frac{1}{n}-\sum\limits_{n=1}^{+\infty}\frac{1}{n+2}\right)=\]\[=3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots-\frac{1}{3}-\frac{1}{4}-\cdots\right)=3\left(1+\frac{1}{2}\right)=4.5\]