Find $\lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2}$

Question

I have a problem with finding the following limit. I suspect that it should be easy, but really I don't have a clue. $$ \lim_{x\to0}\sum_{n=1}^{\infty}\frac{\sin x}{4+n^2x^2} $$

Answer 1

Note that if $(n-1)x \leqslant u \leqslant nx \leqslant v \leqslant (n+1)x,$ then

$$\frac{1}{4 + v^2} \leqslant \frac{1}{4 +n^2x^2} \leqslant \frac{1}{4 + u^2}, $$

and $$\int_{nx}^{(n+1)x}\frac{dv}{4 + v^2} \leqslant \frac{x}{4 +n^2x^2} \leqslant \int_{(n-1)x}^{nx}\frac{du}{4 + u^2}, $$

Summing we get

$$\int_x^\infty \frac{dv}{4 + v^2} \leqslant \sum_{n=1}^\infty \frac{x}{4 +n^2x^2} \leqslant \int_0^\infty \frac{du}{4 + u^2} .$$

Using the squeeze principle,

$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{x}{4 +n^2x^2} = \int_0^\infty \frac{du}{4 + u^2}.$$

Hence,

$$\lim_{x \to 0}\sum_{n=1}^\infty \frac{\sin x}{4 +n^2x^2} = \lim_{x \to 0} \frac{\sin x}{x} \sum_{n=1}^\infty \frac {x}{4 +n^2x^2} \\ = \int_0^\infty \frac{du}{4 + u^2}$$

Answer 2

Without justifying any single detail, I propose the following solution. I introduce the evocative symbol $$ \Delta t:=x $$ and define the sequence $t_n=n\Delta t$. Then $$ \lim_{x\to0}\sum_{n=1}^{\infty}\frac{sinx}{4+n^2x^2}=\lim_{x\to0}\sum_{n=1}^{\infty}\frac{x}{4+n^2x^2}=\lim_{\Delta t\to0}\sum_{t_n\in \Delta t\cdot \mathbb N}\frac{\Delta t}{4+t_n^2}=\int_0^\infty\frac{dt}{4+t^2}=\frac{\pi}{4} $$