The problem:

Show that $\sum_{n=1}^\infty \frac 1n\,\sin\left(\Bigl(n-\frac 12\Bigr)\pi+\frac xn\right)$ converges uniformly and is continuously differentiable for $x \in [-a,a]$ with $a>0$.

Possible theorem involved:

(1) Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a,b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b]$. If $\{f'_n\}$ converges uniformly on $[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$, to a function $f$, and $$f'(x) = \lim_{n \rightarrow \infty} f'_n(x) \hspace 1em (a \leq x \leq b)$$

(2) If $\{f_n\}$ is a sequence of continuous functions on $E$, and if $f_n \to f$ uniformly on $E$, then $f$ is continuous on $E$.


I rewrite the series using trigonometry as: $\sum_{n=1}^\infty \frac{1}{n}\sin\left(\left(n - \frac{1}{2}\right)\pi + \frac{x}{n}\right) = \sum_{n=1}^\infty \frac{1}{n} (-1)^{n+1}\cos \left(\frac{x}{n}\right)$. Let $f_n(x) = \frac{1}{n} (-1)^{n+1}\cos \left(\frac{x}{n}\right)$, then I can differentiate this to get $f'_n(x) = \frac{(-1)^n}{n^2} \sin \left(\frac{x}{n}\right)$.

Consider $\sum f'_n(x)$, this series converges uniformly by comparison with $\sum \frac{1}{n^2}$ (Weierstrass M test) on $[-a,a]$. By theorem (1) mentioned above, $\sum f_n$ also converges uniformly on $[-a,a]$.

Apply theorem (2) to obtain the continuous of $f'_n(x)$. Thus this implies the continuous differentiability of$\sum f_n$.

I'd really appreciate if someone can help me strengthen my proof.

  • 1
    $\begingroup$ You posted the same problem 3 hours ago. You deleted it and now you post it again. Please don't do that. If you wish to alter the question, just edit it. $\endgroup$ – zhw. Nov 20 '15 at 5:21
  • $\begingroup$ Thanks for comment. I posted a part of the question previously, but now I "finished" the problem and hope that someone could check it. I will not do it again in the future. $\endgroup$ – Paichu Nov 20 '15 at 5:24
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    $\begingroup$ I think $\sum f_n$ converges uniformly is not use M test. When $n > a$, then $\cos(x/n)/n $ is $\rightarrow 0$ monotonously. Then use Leibniz Test. $\endgroup$ – Laura Nov 20 '15 at 6:25
  • $\begingroup$ Thank you. That is useful also. Although, I used the theorem stated in the problem to show that $f_n$ converges uniformly. $\endgroup$ – Paichu Nov 20 '15 at 6:38

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