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given is the following series

$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$$

And I need to find its value.

How can I start finding it?

Thanks for all

does the Telescop-Summing work here as well?:

$\sum_{n=1}^\infty \frac{1}{4n^2-1} $ now: $\frac{1}{4n^2-1} = \frac{1}{2} * \frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)} = \frac{1}{2} * ( \frac{1}{2n-1} - \frac{1}{2n+1})$

Now I have to "add the sum": $\sum_{n=1}^\infty \frac{1}{4n^2-1} = \frac{1}{2}* [ \sum_{n=1}^\infty \frac{1}{2n-1} - \sum_{n=1}^\infty \frac{1}{2n+1}] = \frac{1}{2} - \frac{1}{4n+2} $ And than for $n \to \infty$ it is $\frac{1}{2}$ ??

user642796
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Vazrael
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2 Answers2

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HINT: $$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}=\frac1{n^2}-\frac1{(n+1)^2}$$

Can you recognize Telescoping Sum?

lab bhattacharjee
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  • Hi, that's a huge hint, thank you very much! :) If I am right, than using the 'Telescoping Sum' in the end I get: $1 - \frac{1}{(n+1)^2}$ and now I look at $ n \to \infty $ and I get: The Value is 1. – Vazrael May 21 '13 at 13:47
  • @KevinLiebing, welcome. You are right. – lab bhattacharjee May 21 '13 at 14:51
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split it up the numerator to 2n + 2 - 1 then factor to 2(n+1) -1. By spliting the fraction you can cancel terms. This should allow you to see this is a finite result.

Tim
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  • Are you suggesting that we go $\frac{2n+1}{n^2(n+1)^2}=\frac{2(n+1)-1}{n^2(n+1)^2}=\frac{2-1}{n^2(n+1)}=\frac{1}{n^2(n+1)}$?...! – user1729 May 21 '13 at 10:52