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

Since $\frac{k}{n^2+k^2}\leq \frac{k}{k^2+k^2}=\frac{1}{2k}$, then $\sum_{k=1}^n \frac{k}{n^2+k^2}\leq \sum_{k=1}^n \frac{1}{2k}=\frac{1}{2} \sum_{k=1}^n\frac{1}{k}$.

Now we send $n$ to infinity, then since $\sum_{k=1}^\infty \frac{1}{k}$ is harmonic, $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$ doesn't exist.

I wonder if my thinking is right.

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marked as duplicate by YuiTo Cheng, max_zorn, John Omielan, José Carlos Santos sequences-and-series Jun 28 at 21:02

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  • $\begingroup$ @BenjaminDickman Sorry, the previous title is auto-saved. I have corrected it. $\endgroup$ – SHBaoS Mar 28 '17 at 3:33
  • $\begingroup$ All you have shown is that the sum is bounded above by the harmonic series, which you know diverges to infinity. That does not mean that the limit of the sum under consideration doesn't exist. $\endgroup$ – Mattos Mar 28 '17 at 3:35
  • $\begingroup$ I deleted my previous comment, but here is a new one: It seems you have shown your series is bounded above by a series that diverges to infinity. But this is not enough to conclude anything about your series... $\endgroup$ – Benjamin Dickman Mar 28 '17 at 3:35
  • $\begingroup$ If $5 < \infty$ is $5$ infinite? $\endgroup$ – Unit Mar 28 '17 at 3:36
  • $\begingroup$ Thanks for all of you. I think I've made a silly mistake. I will ponder on it for a while. $\endgroup$ – SHBaoS Mar 28 '17 at 3:39
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You can do this as follows,

$$\lim_{n\rightarrow\infty}\sum_{k=1}^n \frac{k}{n^2+ k^2}=\lim_{n\rightarrow\infty}\sum_{k=1}^n \frac{k}{n^2(1+(k/n)^2)}= \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{(1+(k/n)^2)}$$ This corresponds to a certain integration. Calculate that.

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  • $\begingroup$ Do you the integration is $\int \frac{x}{1+x^2}\,dx$? $\endgroup$ – SHBaoS Mar 28 '17 at 4:01
  • $\begingroup$ @SHBaoS: Yes exaclty, with the lower limit $0$ and the upper limit $1$. $\endgroup$ – Parish Mar 28 '17 at 4:06
  • $\begingroup$ But what about the $\frac{1}{n}$ outside the summation? The integral is $\frac{1}{2}ln(2)$, but if $n$ goes to $\infty$, then shouldn't $\frac{1}{n}$ go to zero? $\endgroup$ – SHBaoS Mar 28 '17 at 4:30
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    $\begingroup$ @SHBaoS: Do you know what is Riemann integration. Can you recall how do we get from such limit sum to an integral, I mean how do we prove that the limit sum is an integral. You may find your answer there. You may also think intuitively as the integral representing the area under a curve, and as the 'width' $1/n$ shrinks to $0$ you approach the area ever so closer. $\endgroup$ – Parish Mar 28 '17 at 4:37
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Just for the fun.

Consider $$S_n= \sum_{k=1}^n \frac{k}{n^2+k^2}=\sum_{k=1}^n \frac{k}{(n^2+ik)(n-ik)}=\frac i2 \sum_{k=1}^n\left(\frac{1}{n+i k}-\frac{1}{n-i k} \right)$$ Using generalized harmonic numbers $$S_n=\frac{1}{2} \left(-H_{-i n}-H_{i n}+H_{(1-i) n}+H_{(1+i) n}\right)$$ Using asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ and applying to each term, after simplifications, you should end with $$S_n=\frac{\log (2)}{2}+\frac{1}{4 n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.

Use it with $n=10$ which gives $$S_{10}=\frac{2892380100711541}{7801656832544900}\approx 0.370739$$ while the expansion gives $$S_{10}\approx\frac{\log (2)}{2}+\frac{193}{8400}\approx 0.369550$$

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