Let $D(x):\mathbb{R}\rightarrow\mathbb{R}$ be: $$ D(x)=\sum^\infty_{k=1}\frac{1}{k!}\sin((k+1)!x)$$ Prove that $D(x)$ is nowhere differentiable.
What I've done is that I supposed that exist a $x\in \mathbb{R}$ where $D(x)$ is differentiables, so: $$\lim_{h\rightarrow0} \frac{D(x+h)-D(x)}{h}=a \Rightarrow$$ $$\lim_{h\rightarrow0} \frac{\sum^\infty_{k=1}\frac{1}{k!}\sin[(k+1)!(x+h)]-\sum^\infty_{k=1}\frac{1}{k!}\sin((k+1)!x)}{h}=$$ $$\lim_{h\rightarrow0} \frac{\sum^\infty_{k=1}\frac{1}{k!}[\sin[(k+1)!(x+h)]-\sin((k+1)!x)]}{h}=$$ $$\sum^\infty_{k=1}\frac{1}{k!}\lim_{h\rightarrow0}\frac{ 2\sin(\frac{(k+1)!(x+h)-(k+1)!x}{2})\cos(\frac{(k+1)!(x+h)+(k+1)!x}{2})}{h}=$$ $$\sum^\infty_{k=1}\frac{1}{k!}\lim_{h\rightarrow0}\frac{ 2\sin(\frac{(k+1)!h}{2})\cos(\frac{(k+1)!(2x+h)}{2})}{h}=\frac{0} {0} $$ Using L'hopital: $$ \sum^\infty_{k=1}\frac{1}{k!}\lim_{h\rightarrow0}\frac{\cos(\frac{(k+1)!h}{2})(\frac{(k+1)!}{2})\cos(\frac{(k+1)!(2x+h)}{2})-\sin(\frac{(k+1)!h}{2})(\frac{(k+1)!}{2})\sin(\frac{(k+1)!(2x+h)}{2})}{1} =$$ $$ \sum^\infty_{k=1}\frac{1}{k!}\lim_{h\rightarrow0}[(\frac{(k+1)!}{2})\cos(\frac{(k+1)!(2x+h)}{2})] =$$ $$ \sum^\infty_{k=1}\frac{1}{k!}(\frac{(k+1)!}{2})\cos((k+1)!(x)) = \sum^\infty_{k=1}\frac{k+1}{2}\cos((k+1)!(x)) $$ I don't know how to continue.