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.

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 $h\to0$.
