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

Question

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!

Answer 1

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

Answer 2

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 $x$.