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How do I evaluate this?

$$\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$

I got concerned for that, I've tried make it integral for Riemann but it still undone.

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marked as duplicate by YuiTo Cheng, DMcMor, max_zorn, José Carlos Santos sequences-and-series Jun 28 at 20:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$\sum_{k=1}^{n}\frac{k}{k^{2}+n^{2}}=\frac{1}{n}\sum_{k=1}^{n}\frac{k/n}{\frac{k^{2}}{n^{2}}+1}$$ Which is a Riemann sum for the function $f(x)=\frac{x}{1+x^{2}}$ over $[0,1]$. Your sum therefore tends to $$\int_{0}^{1}\frac{x}{1+x^{2}}dx$$ Can you go on from here?

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  • $\begingroup$ Thank you so much Daniel, yeah i can. $\endgroup$ – Reza Habibi Sep 22 '15 at 12:23
  • $\begingroup$ Both answers come with relevant clarifications on your problem. It would only be fair and in the spirit of the community to accept one $\endgroup$ – Victor Sep 22 '15 at 13:28
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Notice, $$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$ $$=\lim_{n\to \infty}\sum_{k=1}^{n}\frac{\left(\frac{k}{n}\right)\frac{1}{n}}{\left(\frac{k}{n}\right)^2+1}$$ Let $\frac{k}{n}=u\implies \lim_{n\to \infty}\frac{1}{n}=du\to 0$

then we have $$\text{upper bound of u}=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower bound of u}=\lim_{n\to \infty}\frac{k}{n}=\lim_{n\to \infty}\frac{1}{n}=0$$ Changing summation into integration with proper limits $$\int_{0}^{1}\frac{u\ du}{u^2+1}$$ $$=\frac{1}{2}\int_{0}^{1}\frac{(2u)\ du}{u^2+1}=\frac{1}{2}\int_{0}^{1}\frac{d(u^2)}{u^2+1}$$ $$=\frac{1}{2}[\ln|u^2+1|]_{0}^{1}$$ $$=\frac{1}{2}[\ln|1+1|-\ln|0+1|]$$$$=\color{red}{\frac{1}{2}\ln 2}$$

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  • $\begingroup$ Which part of this is not covered by the previous answer? $\endgroup$ – Did Sep 22 '15 at 14:35
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Notice that $$\frac{k}{k^2+n^2}$$ is an increasing function on $k \in (0,n)$ as derivative is positive in this range.

Therefore $$\int_0^{n-1}\frac{k}{k^2+n^2}\leq\sum_1^{n-1}\frac{k}{k^2+n^2}\leq\int_1^{n}\frac{k}{k^2+n^2}$$.

Note that $$\int_0^{n-1}\frac{k}{k^2+n^2}=(1/2) \log \frac{2n^2-2n+1}{n^2}$$

So $$\lim_{n \to \infty}\int_0^{n-1}\frac{k}{k^2+n^2}=\lim_{n \to \infty}(1/2) \log \frac{2n^2-2n+1}{n^2}= 1/2 \log2$$.

Also, $$\int_1^{n}\frac{k}{k^2+n^2}=(1/2) \log \frac{2n^2}{n^2+1}$$

So $$\lim_{n \to \infty}\int_1^{n}\frac{k}{k^2+n^2}=\lim_{n \to \infty}(1/2) \log \frac{2n^2}{n^2+1}= 1/2 \log2$$.

So $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} + \lim_{n\to\infty}\frac{n}{n^2+n^2}=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} + 0$$

But, $$S=\lim_{n\to\infty}\sum_1^{n-1}\frac{k}{k^2+n^2} = (1/2)\log2$$, as $$(1/2)\log2\leq S\leq(1/2)\log2$$

So $$\lim_{n\to\infty} \sum_1^{n}\frac{k}{k^2+n^2}=(1/2)\log2$$

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