# Uniform convergence - Is f continuous and is f differentiable $f(x) = \sum_{n=1}^{\infty}\frac{1}{2n^2-\sin(nx)}$

Let f(x) = $$\sum_{n=1}^{\infty}\frac{1}{2n^2-\sin(nx)}$$ ($$x\in\mathbb{R}$$)

(a) Decide whether f is continuous on R.

(b) Is f differentiable?

Don't even know where to begin with this question, I assume we need to work out if it converges uniformly first but unsure how. I've been practicing with the Weierstrass M-Test for questions such as $$\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}$$ etc but not any of this form. Any help would be great!

## 2 Answers

The partial sums for $$f$$, and the deriatives of the partial sums, converge absolutely and uniformly on $$\mathbb{R}$$. Indeed, if $$f_{N}$$ denotes the $$N$$-th partial sum, note $$f_{N}(x) = \sum_{n =1}^{N} \frac{1}{2n^2-\sin(nx)}, f_{N}'(x)= \sum_{n =1}^{N} \frac{n \cos(nx)}{(2n^2-\sin(nx))^2}$$

and we have that $$\left|\frac{1}{2n^2 - \sin(nx)}\right| \leq \frac{1}{2n^2-1}, \ \ \ \left|\frac{n \cos(nx)}{(2n^2-\sin(nx))^2}\right| \leq \frac{n}{(2n^2-1)^2}$$

• beat me to it narrowly! – George Coote May 9 at 21:49

Noting that:

$$\frac 1 {2n^2 - \sin (nx)} \le \frac 1 {2n^2 - 1}$$

since $$-1 \le \sin nx \le 1$$ ($$x$$, $$n$$ are real) should help you with the Weierstrass M-test.

For the second part, note that we can examine the uniform convergence of:

$$\sum_{n = 1}^\infty \frac \partial {\partial x} \left(\frac 1 {2n^2 - \sin(nx)}\right)$$

to give a sufficient condition for $$f$$ being differentiable and being equal to its "termwise derivative", noting that each term is differentiable in $$x$$.