# Limit of a peculiar sum $\sum_{n=1}^{\infty}\frac{x}{1+x^2n^2}$

Find the limit (is exists) of $$\lim_{x\longrightarrow 0^+}\sum_{n=1}^{\infty}\frac{x}{1+x^2n^2}.$$ I tried turning it into Riemann sum, but to no avail as of yet.

-
$\displaystyle\sum_{n=-\infty}^\infty\frac1{1+x^2n^2}=\frac\pi x\coth\frac\pi x$ –  Lucian Jun 16 '14 at 17:00

You can indeed handle it as a Riemann sum for $\int_0^\infty(1+x^2)^{-1}\,dx=\pi/2$: $$\int_0^\infty\frac{dx}{1+x^2}-\sum_{n=1}^\infty \frac h{1+h^2n^2} =\sum_{n=1}^\infty\int_{(n-1)h}^{nh}\Bigl(\frac1{1+x^2}-\frac1{1+n^2h^2}\Bigr)\,dx\ge0$$ while $$\int_h^\infty\frac{dx}{1+x^2}-\sum_{n=1}^\infty \frac h{1+h^2n^2} =\sum_{n=1}^\infty\int_{nh}^{(n+1)h}\Bigl(\frac1{1+x^2}-\frac1{1+n^2h^2}\Bigr)\,dx\le0$$ so $$\int_h^\infty\frac{dx}{1+x^2}\le\sum_{n=1}^\infty \frac h{1+h^2n^2}\le\int_0^\infty\frac{dx}{1+x^2}$$ and now you can let $h\to0$.

-
Ok, I accepted you answer, but could you please elaborate on how do you take the integral under the sum (I get the first part, but don't know how this sum is actually a sum of integrals)? –  Jules Jun 16 '14 at 12:32
Just use $\int_0^\infty(1+x^2)^{-1}\,dx=\lim_{N\to\infty}\int_0^{Nh}(1+x^2)^{-1}\,dx$, then split this up as a sum of $N$ integrals, and note that $\sum_{n=1}^\infty=\lim_{N\to\infty}\sum_{n=1}^N$. –  Harald Hanche-Olsen Jun 16 '14 at 12:35
I know, but you say $\sum_{n=1}^\infty \frac h{1+h^2n^2}=\sum_{n=1}^\infty\int_{(n-1)h}^{nh}\frac1{1+n^2h^2}\ dx$ and I'm trying to see why. –  Jules Jun 16 '14 at 12:40
I finally figured it out, you're right it's not that complicated ;) –  Jules Jun 16 '14 at 13:39
Good. Note that this is quite a common trick for estimating integrals. You'll likely see it again in different contexts. –  Harald Hanche-Olsen Jun 16 '14 at 13:43

You can actually intepret it as a Riemann sum in the following way

$$\lim_{x\rightarrow 0+}\sum_{n=1}^{\infty}\frac{x}{1+(xn)^{2}}=\lim_{m\rightarrow\infty}\sum_{n=1}^{\infty}\frac{1}{m}\frac{1}{1+(\frac{n}{m})^{2}}=\int_{0}^{\infty}\frac{dt}{1+t^{2}}=\frac{\pi}{2}$$

-
I tried doing it that way! I can just set integration bounds like this? Shouldn't it be like $\lim_{m\rightarrow\infty}\sum_{n=1}^{m}\frac{1}{m}\frac{1}{1+(\frac{n}{m})^{2}}‌​$ ? –  Jules Jun 16 '14 at 12:06
I didn't see this answer until now, while I was typing away at my own. Anyhow, my answer has more detail which should make it clear. I'll +1 this in the spirit of good sportsmanship, though. –  Harald Hanche-Olsen Jun 16 '14 at 12:21
Since this is an improper integral, I think you may need to take a little more care showing that you have the correct limit of Riemann sums. I find $\int_{0}^{\infty}\frac{dx}{1+x^2}=\lim_{L\to\infty}\int_{0}^{L}\frac{dx}{1+x^2}‌​=\lim_{L\to\infty}\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+L^2k^2}$, but I'm not completely sure how to connect this double limit to your Riemann sum. –  David H Jun 16 '14 at 12:22
@HaraldHanche-Olsen Ah, and I posted my comment before I saw your answer and comment. It looks your answer handles the issue I was worried about. :) –  David H Jun 16 '14 at 12:24

Hint: the series itself has the closed form

$$\frac12(\pi\coth{\frac{\pi}{x}}-x)$$

-
@Shaktal it's the negative of that. Thanks for alerting me to the typo. –  David H Jun 16 '14 at 11:34
Where does it come from? –  Jules Jun 16 '14 at 11:42
@Jules See this question for a proof: math.stackexchange.com/questions/680825/… –  David H Jun 16 '14 at 11:49
Another approach may be to use the product formula for the sine function – se formula (23) on the mathworld article on the sine function – take the logarithm of that, then differentiate. (And then replace $x$ by $ix$.) –  Harald Hanche-Olsen Jun 16 '14 at 11:53