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michael h's question at Yahoo! Answers (Sum of a series)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello michael h,

We have $(n+1)^2-n^2=2n+1$, so $$\dfrac{2n+1}{n^2(n+1)^2}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}$$ Using that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2},\;\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{(n+1)^2}$ are both convergent: $$\displaystyle\sum_{n=1}^{\infty}\dfrac{2n+1}{n^2(n+1)^2}=\displaystyle\sum_{n=1}^{\infty}\left( \dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}\right)=\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{(n+1)^2}=\\\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\left(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\dfrac{1}{1^2}\right)=\left(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\sum_{n=1}^{\infty}\dfrac{1}{n^2}\right)+1=0+1=1$$