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!


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

| cite | improve this answer | |
  • $\begingroup$ beat me to it narrowly! $\endgroup$ – 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 $x$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.