MHB Michael h's question at Yahoo Answers (Sum of a series)

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The discussion centers on finding the sum of the series represented by the expression (2n+1)/(n^2(n+1)^2). The transformation of the expression reveals that it can be rewritten as a difference of two convergent series: 1/n^2 and 1/(n+1)^2. By applying the properties of these series, the sum converges to 1. The calculations demonstrate that the series converges effectively, leading to a clear conclusion. The final result of the sum is 1.
Fernando Revilla
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Here is the question:

(2n+1)/(n^2(n+1)^2)

Here is a link to the question:

Find the sum of the series? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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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: $$\displaystyle\sum_{n=1}^{\infty}\dfrac{2n+1}{n^2(n+1)^2}=\displaystyle\sum_{n=1}^{\infty}\left( \dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}\right)=\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{(n+1)^2}=\\\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\left(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\dfrac{1}{1^2}\right)=\left(\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}-\sum_{n=1}^{\infty}\dfrac{1}{n^2}\right)+1=0+1=1$$
 
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