I got asked this question in a group, and solved it the following way:

$$\lim\limits_{n\to\infty}\left( \int\limits_1^{n+1}\frac{x}{x^2+n^2}\space dx\leq \sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}\leq \int\limits_1^n \frac{x}{x^2+n^2}\space dx + f(1)\right) $$

$$\frac{\ln2}{2}\leq \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2}\leq\frac{\ln2}{2}$$


$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{k^2+n^2} = \frac{\ln2}{2}$$

The guy who asked me the question said that the answer key said $\frac{\ln\left(\frac{5}{2}\right)}{2}$ but also said that it might be incorrect, so I don't have the answer.

What I want to ask is that how come this sum is not equal to $0$? We were able to use the squeeze theorem because the limit for $f(1)$ goes to zero. Every other value after $f(1)$ also goes to $0$ (even faster?) and that had me thinking about how this summation can be equal to such a value.

Can anyone please explain how this happens? And it would be great if you could also verify or correct my solution.

  • $\begingroup$ The sum $\sum_{k=1}^n (1)=n$. So we have $\sum_{k=1}^n \frac1n=1$. And hence $\lim{n\to\infty}\sum_{k=1}^n \frac1n=1$ even though each term of the summand $\frac1n$ goes to $0$. $\endgroup$
    – Mark Viola
    Nov 7 '20 at 15:33
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    $\begingroup$ Yes, your solution is correct and a nice application of the integral test for convergence, though the sum in question is perfectly set-up for applying Riemann sum as given in Z Ahmed's answer. Also, regarding your question, the reason could be seen as "sum of infinitesimals need not be infinitesimal"; for example, consider $f(k)=\frac 1{n+k}$ which all go to $0$ as $n\to\infty$ and $f$ decays faster as $k$ increases when $n\to\infty$, but the sum $\sum_{k\geq 1} f(k)$ is a tail of the harmonic series, which diverges. $\endgroup$ Nov 7 '20 at 15:35
  • $\begingroup$ You should go and double check the bound of the sum in original question. The number $\frac12\log(\frac52)$ corresponds to the limit $\lim_{n\to\infty}\sum\limits_{k=\color{red}{n+1}}^{\color{red}{2n}}\frac{k}{k^2+n^2}$ $\endgroup$ Nov 7 '20 at 15:48
  • $\begingroup$ @achillehui. In my edit, I computed $\sum_{k=a n +b}^{c n+d}\frac k {k^2+n^2}$. If I am not mistaken, the limit does not depend on $b$ or $d$. Do you mind to check ? Thanks & cheers :-) $\endgroup$ Nov 8 '20 at 2:28

If you write $$\frac k {k^2+n^2}=\frac k {(k+i n)(k-in)}=\frac i 2 \left(\frac 1 {k+i n}-\frac 1 {k-i n} \right)$$ $$S_n=\sum_{k=1}^n\frac k {k^2+n^2}=\frac{1}{2} \left(H_{(1-i) n}+H_{(1+i) n}-H_{-i n}-H_{i n}\right)$$ Using asymptotics $$H_p=\gamma+\log (p)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ $$S_n=\frac{\log (2)}{2}+\frac{1}{4 n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$


After @achille hui's comment, I consider the most general case of $$\Sigma=\sum_{k=a n +b}^{c n+d}\frac k {k^2+n^2}$$ Using the same method, I obtained as asymptotics $$\Sigma=\frac{1}{2} \log \left(\frac{c^2+1}{a^2+1}\right)+\frac{2 \left(a^2+1\right) c d-2 a b \left(c^2+1\right)+(a+c) (a c+1)}{2\left(a^2+1\right) \left(c^2+1\right) n}+O\left(\frac{1}{n^2}\right)$$ and then the limit only depends on $a$ and $c$ (nothing to do with $b$ and $d$ which only define how is approached the limit).

  • $\begingroup$ I didn't check the aysmptotics but as long as $b$ and $d$ fixed, it won't contribute to the final limit b/c $\sum_{k=cn+1}^{cn+b} \frac{k}{k^2+n^2} \sim b\frac{cn}{(cn)^2 + n^2} = \frac{bc}{n(1+c^2)}$, same thing happens to $d$. $\endgroup$ Nov 8 '20 at 3:29

$$S=\lim_{n \to \infty} \frac{k}{n^2+k^2}=\lim_{n \to \infty} \sum_{k=0}^{n}\frac{1}{n} \frac{k/n}{1+(k/n)^2}=\int_{0}^{1}\frac{x}{1+x^2} dx=\frac{1}{2} \ln(1+x^2)|_{0}^{1}$$ $$=\frac{1}{2} \ln 2$$

  • $\begingroup$ Could you please explain how we get from summation to integral? $\endgroup$
    – Lars Smith
    Nov 7 '20 at 16:34
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    $\begingroup$ @LarsSmith: It's Riemann sum which I mentioned in the comments. We have, $$\frac 1n\sum_{k=an}^{bn} f(k/n)\to\int_a^b f(x)~\mathrm dx$$ as $n\to\infty$ where $a\leq b$ are non-negative integers. It also holds if you add or subtract finitely many elements of the sum, say sum from $an$ to $bn-1$ or $an+1$ to $bn$, etc $\endgroup$ Nov 7 '20 at 16:38

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