- Thread starter
- #1

- Jan 29, 2012

- 661

Here is a link to the question:(2n+1)/(n^2(n+1)^2)

Find the sum of the series? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter Fernando Revilla
- Start date

- Thread starter
- #1

- Jan 29, 2012

- 661

Here is a link to the question:(2n+1)/(n^2(n+1)^2)

Find the sum of the series? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter
- #2

- Jan 29, 2012

- 661

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: $$\displaystyle\sum_{n=1}^{\infty}\dfrac{2n+1}{n^2(n+1)^2}=\displaystyle\sum_{n=1}^{\infty}\left( \dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}\right)=\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{(n+1)^2}=\\\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\left(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\dfrac{1}{1^2}\right)=\left(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\sum_{n=1}^{\infty}\dfrac{1}{n^2}\right)+1=0+1=1$$