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I guess it has something to do with Riemann sums but this is new for me.

$\displaystyle\lim \limits_{n \to \infty}\sum \limits_{k=n}^{2n}\frac{k}{k^2+n^2}$

How do I start?

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marked as duplicate by YuiTo Cheng, Leucippus, mrtaurho, Hayk, Thomas Shelby Jun 29 at 11:39

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Like this: $$ \sum_{k=n}^{2n}\frac{k}{k^2+n^2}=\frac1n\sum_{i=0}^{n}\frac{1+\frac{i}n}{\left(1+\frac{i}n\right)^2+1}. $$

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$$S_n = \sum_{k=n}^{2n} \dfrac{k}{k^2+n^2} = \sum_{k=n}^{2n} \dfrac{k/n}{(k/n)^2+1} \dfrac1n$$ Let $x_{k-n+1} = k/n$, we then have $$S_n = \sum_{k=1}^n \dfrac{x_k}{1+x_k^2} \Delta x_k$$ You should be able to relate the above sum to $$\int_1^2 \dfrac{x}{1+x^2} dx$$

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  • $\begingroup$ thank you but is the result of the integral enough to argument that the series converges? $\endgroup$ – PaulH Apr 24 '13 at 21:00
  • $\begingroup$ As long as the integral exists, yes: it is. $\endgroup$ – DonAntonio Apr 24 '13 at 21:10

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