How to prove that $ f(x) = \sum_{k=1}^\infty \frac{\sin((k + 1)!\;x )}{k!}$ is nowhere differentiable This function is continuous, it follows by M-Weierstrass Test. But proving non-differentiability, I think it's too hard. Does someone know how can I prove this? Or at least have a paper with the proof?
The function is 
$$
f(x) = \sum_{k=1}^\infty \frac{\sin((k + 1)!\;x )}{k!}$$
Thanks!
 A: (Edited: handwaving replaced by rigor)
For conciseness, define the helper functions $\gamma_k(x)=\sin((k+1)!x)$. Then $f(x)=\sum_k \frac{\gamma_k(x)}{k!}$.
Fix an arbitrary $x\in\mathbb R$. We will construct a sequence $(x_n)_n$ such that
$$\lim_{n\to\infty} x_n = x \quad\land\quad \lim_{n\to\infty} \left|\frac{f(x_n)-f(x)}{x_n-x}\right| = \infty$$
Such a sequence will directly imply that $f$ is not differentiable at $x$.
Let $x'_n$ be the largest number less than $x$ such that $|\gamma_n(x'_n)-\gamma_n(x)|=1$. Let $x''_n$ be the smallest number larger than $x$ such that $\gamma_n(x''_n)=\gamma_n(x'_n)$. One of these, to be determined later, will become our $x_n$.
No matter which of these two choices of $x_n$ we have $|x_n-x|<\frac{2\pi}{(n+1)!}$ so $\lim x_n=x$.
To estimate the difference quotient, write
$$f(x) = \underbrace{\sum_{k=1}^{n-1}\frac{\gamma_k(x)}{k!}}_{p(x)}+
\underbrace{\frac{\gamma_n(x)}{n!}}_{q(x)}+
\underbrace{\sum_{k=n+1}^{\infty} \frac{\gamma_k(x)}{k!}}_{r(x)}$$
and so,
$$\underbrace{f(x_n)-f(x)}_{\Delta f} = \underbrace{p(x_n)-p(x)}_{\Delta p} +
\underbrace{q(x_n)-q(x)}_{\Delta q} +
\underbrace{r(x_n)-r(x)}_{\Delta r}$$
Of these, by construction of $x_n$ we have $|\Delta q| = \frac{1}{n!}$.
Also, $r(x)$ is globally bounded by the remainder term in the series $\sum 1/n! = e$, which by Taylor's theorem is at most $\frac{e}{(n+1)!}$. So $|\Delta r| \le \frac{2e}{(n+1)!}$.
$\Delta p$ is not dealt with as easily. In some cases it may be numerically larger than $\Delta q$, ruining a simple triange-equality based estimate. But it can be tamed by a case analysis:


*

*If $p$ is strictly monotonic on $[x'_n, x''_n]$, then $p(x'_n)-p(x)$ and $p(x''_n)-p(x)$ will have opposite signs. Since $q(x'_n)=q(x''_n)$, we can choose $x_n$ such that $\Delta p$ and $\Delta q$ has the same sign. Therefore $|\Delta p+\Delta q|\ge|\Delta q|=\frac{1}{n!}$.

*Otherwise, $p$ has an extremum between $x'_n$ and $x''_n$; select $x_n$ such that the extremum is between $x$ and $x_n$. Because $p$ is a finite sum of $C^\infty$ functions, we can bound its second derivative separately for each of its terms:
$$\forall t: |p''(t)| \le \sum_{k=1}^{n-1}\left|\frac{\gamma''_k(t)}{k!}\right| \le
\sum_{k=1}^{n-1}\frac{(k+1)!^2}{k!} \le
\sum_{k=1}^{n-1} (k+1)!(k+1) \le 2n!n $$
Therefore the maximal variation of $p$ in an interval of length $\le\frac{2\pi}{(n+1)!}$ that contains a stationary point must be $\left(\frac{2\pi}{(n+1)!}\right)^2 2n!n = \frac{8\pi^2n}{(n+1)^2}\frac{1}{n!}$. The $\frac{8\pi^2n}{(n+1)^2}$ factor is less than $1/2$ for $n>16\pi^2$, so for large enough $n$ we have $|\Delta p+\Delta q|\ge \frac{1}{2n!}$.
Thus, for large $n$ we always have
$$|\Delta f| \ge \frac{1}{2n!} - \frac{2e}{(n+1)!} = \frac{1}{n!}\left(\frac{1}{2}-\frac{2e}{n+1}\right)$$
and therefore
$$\left|\frac{f(x_n)-f(x)}{x_n-x}\right| \ge \frac{(n+1)!}{2\pi}|\Delta f| \ge \frac{n+1}{2\pi}\left(\frac{1}{2}-\frac{2e}{n+1}\right) = \frac{n+1}{4\pi}-\frac{e}{\pi} \to \infty$$
as promised.
