# Does the limit of $f(x) = \sum\limits_{n=1}^{+\infty} \frac{(-1)^n \sin \frac{x}{n}}{n}$ exist?

Consider the function $f: [0; +\infty] \rightarrow \mathbb{R}$ given by the formula $$f(x) = \sum\limits_{n=1}^{+\infty} \frac{(-1)^n \sin \frac{x}{n}}{n}$$ Does $\lim\limits_{x\rightarrow 0} f(x)$ exist? If yes, find it's value. Is $f$ differantiable? If yes, check if $f'(0) > 0$.

Any ideas how to do that? I assume this is about functions series, it's pointwise convergent I think, does that mean that this limit exists?

Thanks a lot for your help!

• Yes, it's zero. – Von Neumann Aug 30 '16 at 17:31
• What is this $[0,\infty]$ business? – zhw. Sep 8 '16 at 21:10

Hints: Note that $|\sin u| \leq |u|$ for the first two questions. For differentiability calculate the term-wise derivative and use a theorem for when you are allowed to do so on a series. You may calculate $f'(0)$ explicitly.

First, before you even define $f(x),$ you need to show the series converges at least pointwise. Let $f_n(x) = (-1)^n \sin (x/n)/n.$ Let $a>0.$ Then for $x\in [-a,a],$ $|f_n(x)| \le |x/n|/n\le a/n^2.$ Since $\sum_n a/n^2<\infty,$ the Weierstrass M test shows $\sum f_n(x)$ converges uniformly on every $[-a,a].$ So now we have earned the right to define

$$f(x) = \sum_{n=1}^{\infty}f_n(x), x\in \mathbb R.$$

Since the convergence is uniform in $[-1,1],$ we have

$$\lim_{x\to 0} f(x ) = \sum_{n=1}^{\infty}\lim_{x\to 0}\, f_n(x) = \sum_n 0 = 0.$$

For differentiability, since we already know $\sum_n f_n(x)$ converges everywhere, all we have to do is check that $\sum_{n=1}^{\infty}f_n'(x)$ converges uniformly. That's easy: $|f_n'(x)| = |(-1)^n \frac{\cos (x/n)}{n^2}| \le 1/n^2,$ everywhere. By Weierstrass M, $\sum_n f_n'(x)$ converges uniformly on $\mathbb R,$ so by the standard differentiability/uniform convergence criterion,

$$f'(x) = \sum_{n=1}^{\infty} \frac{(-1)^n \cos (x/n)}{n^2}, \,\, x\in \mathbb R.$$

Therefore $f'(0) = -1 + 1/2^2 - 1/3^2 + \cdots .$ Now use your alternating series spidey sense: This series converges to a negative value, so $f'(0) < 0.$

$f(x)$ is continuous around $x$ and $\lim\limits_{x\to 0}\frac{\sin(x/n)}{x/n}=1$ for any $n$.

Therefore $f(0)=f(x)|_{x\to 0}=x|_{x\to 0}\sum\limits_{n=1}^\infty (-1)^n \frac{1}{n^2}=0\cdot (-\frac{\pi^2}{12})=0$.

$f'(x)=\sum\limits_{n=1}^\infty(-1)^n\frac{\cos(x/n)}{n^2}$ and exists because of $|f'(x)|\leq \sum\limits_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$.

$f'(0)=\sum\limits_{n=1}^\infty(-1)^n\frac{1}{n^2}=-\frac{\pi^2}{12}<0$.