Uniform convergence of a series with sin

Question

Uniform convergence. I need to use the definition or the Cauchy criterion. The series:

$$\sum_{n=1}^{\infty} \left(\sin\dfrac{1}{1+nx}\right)^2.$$

Also I need to find the radius of convergence of the series. Might you help me with that?

Answer 1

We have non-uniform convergence on a set $S$ where $0$ is a limit point. For example, consider $S = (0,\infty)$. Using the inequality $2x/\pi \leqslant \sin x$ for $0 \leqslant x \leqslant \pi/2$. and $1/(1+nx) \leqslant 1$ we have

$$\frac{4}{\pi^2}\frac{1}{(1+nx)^2} \leqslant \sin ^2\left(\frac{1}{1+nx} \right) ,$$

and

$$\sum_{k=n+1}^{2n} \sin ^2\left(\frac{1}{1+kx}\right) \geqslant n \cdot \sin ^2\left(\frac{1}{1+2nx}\right) \geqslant n \cdot \frac{4}{\pi^2}\frac{1}{(1+2nx)^2}$$

With $x = 1/n$ we see that

$$\sup_{x \in (0,\infty)}\sum_{k=n+1}^{2n} \sin ^2\left(\frac{1}{1+kx}\right) \geqslant \frac{4n}{\pi^2(1 + 2n \cdot \frac{1}{n})}= \frac{4n}{3\pi^2}$$

Since the RHS does not converge to $0$ -- in fact, it diverges to $+\infty$ -- as $n \to \infty$, it follows that the Cauchy criterion for uniform convergence is violated.

On the other hand, can you show that the convergence is uniform for $x \in [\delta, \infty)$ when $\delta > 0$? Also what happens if $x < 0$?