Thanks a lot for your help for the previous question: Let $f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin(\frac{x}{n})$. Where is $f$ defined? Is it continuous? Differentiable? Twice-Differentiable?
$f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin(\frac{x}{n})$$f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin\left(\frac{x}{n}\right)$
So I've fully understood how the function is defined for all $x\in\mathbb{R}$ and continuous (through the concept of the uniform convergence using the Mean Value Theorem and the Weierstrass M-Test), but I'm now stuck at its differentiabiltiy and twice-differentibility.
I first naively assumed something similar to Differentiability of a Uniformly Convergent Series my basic idea is that if a series is uniformly convergent then it's also differentiable, using *Term-by-Term Differentiablity Theorem, but the answer above the posting says that it cannot be generalized.
How should I prove $f(x)$ in my case is differentiable (and/or twice-differentiable as well)?
* Term-by-Term Differentiability Theorem: Let $f_n$ be differentiable funcitons defined on an interval A, and assume $\sum\limits_{n=1}^\infty f'_n(x)$ converges unifomly to a limit $g(x)$ on A. If there exists a point $x_0 \in [a,b]$ where $\sum\limits_{n=1}^\infty f_n(x_0)$ converges, then the series $\sum\limits_{n=1}^\infty f_n(x)$ converges uniformly to a differentiable function $f(x)$ satisfying $f'(x)=g(x)$ on A. In other words, $f(x) = \sum\limits_{n=1}^\infty f_n(x)$ and $f'(x)=\sum\limits_{n=1}^\infty f'_n(x)$
