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

Question

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.

Answer 1

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 $h\to0$.

Answer 2

Hint: the series itself has the closed form

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

Answer 3

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

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