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How can I evaluate $\lim_{n \to \infty} \sum_{k=1}^n \frac{k^2}{n^2 + k^2}$? I tried integrating $\lim_{n\to \infty}\int_1^n \frac{k^2}{n^2 + k^2}dk$ but didn't get a finite answer. How to do this?

I know the answer is less than $\pi/4$ because the sum is less than the area under $1/(1+x^2)$ from $1$ to infinity.

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    $\begingroup$ The limit as written is $+\infty$. For $k \geqslant n/2$, the term is $\geqslant 1/5$, so the sum $> n/10$. $\endgroup$ – Daniel Fischer Dec 22 '13 at 10:52
  • $\begingroup$ @DanielFischer The sum diverges? I calculate it's less than $\pi/4$ by comparing with graph of $1/(1+x^2)$. $\endgroup$ – Dylan B. Dec 22 '13 at 10:53
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    $\begingroup$ Yes. Maybe the term should have been $\dfrac{n}{n^2+k^2}$? Or $\dfrac{k^2}{n(n^2+k^2)}$? $\endgroup$ – Daniel Fischer Dec 22 '13 at 10:54
  • $\begingroup$ @DanielFischer If it is what I have written we can write it as $\lim_{n\to \infty} \sum_{k=1}^n 1/(( n/k)^2 + 1)$. That is less than area under graph from $1$ to infty of $1/(1+x^2)$ $\endgroup$ – Dylan B. Dec 22 '13 at 10:56
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    $\begingroup$ Correct, $$\frac{k}{n^2+k^2} = \frac{1}{n}\cdot \frac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}.$$ $\endgroup$ – Daniel Fischer Dec 22 '13 at 11:04
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If we denote $$S_n=\sum_{k=0}^n \frac{k^2}{n^2 + k^2}=\sum_{k=0}^n \frac{\left(\frac kn\right)^2}{1 + \left(\frac kn\right)^2}$$ then we see that $\frac{S_n}{n}$ is a Riemann sum and $$\frac{S_n}{n}\to\int_0^1\frac{x^2}{1+x^2}dx=1-\int_0^1\frac{1}{1+x^2}dx=1-\arctan1=1-\frac\pi 4$$ It's clear that $\lim_n S_n=+\infty$.

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    $\begingroup$ Nicely Done Sami. $\endgroup$ – mrs Dec 22 '13 at 11:16

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