The limit of two expressions involving infinite series

Question

$$\lim_{h\to 0^+} h \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(nh)^{2}} = \lim_{h\to 0} \frac{1}{h} \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(n)^{2}}$$

I tried turning the left side into integral, but it doesn't seem to help. What is the correct way to approach this question?

Answer 1

One may write $$ \begin{align} \lim_{h\to 0^+} h \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(nh)^{2}}&=\lim_{h\to 0^+} h \left[\lim_{M\to \infty}\sum_{n=1}^M \frac {\sin^{2} (nh)} {(nh)^{2}}\right] \\&=\lim_{h\to 0^+} h \left[\lim_{M\to \infty}\frac1{h^2}\sum_{n=1}^M \frac {\sin^{2} (nh)} {(n)^{2}}\right] \\&=\lim_{h\to 0^+} h\left[\frac1{h^2}\lim_{M\to \infty}\sum_{n=1}^M \frac {\sin^{2} (nh)} {(n)^{2}}\right]\quad (\text{convergence of the series*}) \\&= \lim_{h\to 0} \frac{1}{h} \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(n)^{2}} \end{align} $$ as desired.

$\text{*}$Notice the uniform convergence $$ \left| \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(n)^{2}}\right|\le \sum_{n=1}^\infty \frac {1} {(n)^{2}}<\infty. $$

Answer 2

$$\lim_{h\to 0^+} h \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(nh)^{2}} =\\\lim_{h\to 0^+} h \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {n^2h^{2}} =\\ \text{h is constant here } \\\lim_{h\to 0} \frac{h}{h^2} \sum_{n=1}^\infty \frac {\sin^{2} (nh)} {(n)^{2}}$$

Answer 3

Forget the limit signs at first, since you don't know the limit exists until you prove it does.

Without the limit signs: For any fixed $h\ne 0,$ both infinite series converge (the summands being on the order of $1/n^2$). The term $h^2,$ which is a constant with respect to the summation, can thus be pulled out of the series on the left, and we see that the two expressions are identical. There is not much going on there.

But you have theses limit signs. What are they doing there? Is part of the problem is to show the limit exists?

The limit does exist, but this requires some work. You mentioned that the expression on the left might have something to do with an integral. Indeed it does. It looks like an "infinite Riemann sum" for

$$\int_0^\infty \left (\dfrac{\sin x}{x}\right )^2\, dx = \frac{\pi}{2}.$$

Like I said, it require some justification to see the desired limit equals the above integral. I'll leave it here for now.