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I have to find the limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{2k+1}{k^2(k+1)^2}.$$ I tried to make it into a telescopic series but it doesn't really work out...

$$\lim_{n\to\infty} \sum_{k=1}^n \frac{2k+1}{k^2(k+1)^2}=\sum_{k=1}^n \left(\frac{1-k}{k^2}+\frac1{k+1}-\frac1{(k+1)^2} \right)$$ so that is what I did using telescopic...

I said that:

$$\frac{2k+1}{k^2(k+1)^2}=\frac{Ak+B}{k^2}+\frac C{k+1}+\frac D{(k+1)^2}$$ but now as I look at it.. I guess I should "build up the power" with the ${k^2}$ too, right?

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    $\begingroup$ Show us what you did with the telescoping series. $\endgroup$ – heropup Apr 8 '18 at 17:34
  • $\begingroup$ @heropup in a sec $\endgroup$ – C. Cristi Apr 8 '18 at 17:35
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    $\begingroup$ Hint: $\;\; \dfrac{\color{red}{k^2}+2k+1 \color{red}{-k^2}}{k^2(k+1)^2}$ $\endgroup$ – dxiv Apr 8 '18 at 17:40
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    $\begingroup$ @dxiv Thank you. I was too blind to see... Sorry for asking this trivial question.... I shouldn't, it's not my best day. My apologies. $\endgroup$ – C. Cristi Apr 8 '18 at 17:41
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    $\begingroup$ @C.Cristi Next time you'll see it. That's what practice is for. $\endgroup$ – dxiv Apr 8 '18 at 17:49
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$$\lim_{n\rightarrow\infty}\sum^{n}_{k=1}\bigg[\frac{1}{k^2}-\frac{1}{(k+1)^2}\bigg]$$

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