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Determine if $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ converges uniformly and if it can be differentiated term-by-term.

I think I managed to solve this, but I'm not quite sure about the differentiability. Here's what I did. $$|\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}| \leq \sum_{n = 1}^{\infty}\frac{1}{n^2}, \text{ and this sum converges} \Rightarrow \text{From the Weierstrass test, } \sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2} \text{ converges uniformly.}$$

For differentiation, I need to check that, if $f(x) = \frac{\cos(nx)}{n^2}$, $f'(x)$ is continuous (which it is), and that $\sum_{n=1}^{\infty}f'(x) = - \sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$ is uniformly convergent. To do this, I attempt Dirichlet's test:

$$\frac{1}{n} \text{ is obviously monotone and converges to } 0.$$

$$\sin\frac{x}{2}\sum_{n=1}^{m}\sin(nx) = \sum_{n=1}^{m}\sin\frac{x}{2}\cdot\sin(nx)=\frac{1}{2}\sum_{n=1}^{m}(\cos(x(n-\frac{1}{2}))-\cos(x(n+\frac{1}{2}))) = \frac{1}{2}(\cos(\frac{x}{2}) - \cos(\frac{2m+1}{2})) \Rightarrow |\sum_{n=1}^{\infty}\sin(nx)| = \lim_{m\to\infty}\frac{|\cos(\frac{x}{2}) - \cos(\frac{2m+1}{2})|}{2\cdot\sin\frac{x}{2}} \leq \lim_{m\to\infty}\frac{|\cos(\frac{x}{2})| + |\cos(\frac{2m+1}{2})|}{2\cdot\sin\frac{x}{2}} \leq \frac{1}{\sin\frac{x}{2}}$$

Dirichlet's test's hypothesis is fulfilled, thus $\sum_{n = 1}^{\infty}f'(x)$ is uniformly convergent. Finally, from all these, $\sum_{n=1}^{\infty}f(x)$ can be differentiated term-by-term, such that $\left(\sum_{n=1}^{\infty}f(x)\right)' = \sum_{n=1}^{\infty}f'(x)$.

Is my proof correct? Have I missed anything? Any help is much appreciated!

J__n
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  • Doesn't $\sum\limits_{n=1}^{n}$ look strange? – user58697 Dec 07 '23 at 21:55
  • @user58697 Indeed, I forgot to change the index to k, I edited the post now. – J__n Dec 07 '23 at 22:09
  • The solution is wrong. Note that you didn't get uniform boundedness, as you also have the $\sin(\frac{x}{2})$ factor. (which is equal to $0$ at some points by the way) Actually, the series $\sum\limits_{n=1}^{\infty}\frac{\sin(nx)}{n}$ is not uniformly convergent in $\mathbb{R}$. See here: https://math.stackexchange.com/questions/28830/does-sum-frac-sinnxn-converge-uniformly-for-all-x-in-0-2-pi – Mark Dec 07 '23 at 22:17
  • Here is a similar problem with $\sin$ instead of $\cos$: https://math.stackexchange.com/questions/2383397/differentiability-of-the-sum-of-sum-n-1-infty-frac-sinnxn2 – Mark Dec 07 '23 at 22:23
  • @Mark I see now. I tried to apply to this a method from a previous exercise we were shown, and I ended up butchering it. Thank you for the help! – J__n Dec 07 '23 at 22:25
  • The answer to differentiating is actually yes but you have to treat the cases $x=0$ (and all the periods $2k\pi$) and $x\ne0$ (or any period) differently since the derivative is discontinuous at zero; the second case is easy since the sine series you get converges uniformly on any compact that doesn't contain $0$ or any other period; the zero case just do the limit and split the corresponding series carefully; note that for the analogue problem but with $\sin nx/n^2$ the derivative is infinite at $0$ so that problem is slightly different – Conrad Dec 07 '23 at 23:23

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