Continuity, differentiability and double-differentiability of $\sum_{k=1}^{\infty} \frac{\sin(x/k)}{k}$ and $\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^3}$ I have solved the following exercise except for the last point and I would like to know if I have made any mistakes and I would appreciate any hint about how to complete the last question in part (b) about $g$ being twice-differentiable; thanks.
Consider $f(x)=\sum_{k=1}^{\infty} \frac{\sin(x/k)}{k}$ and $g(x)=\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^3}$.
(a) Where is $f$ defined? Continuous? Differentiable? Twice-differentiable?
(b) Show that $g(x)$ is differentiable and that $g'(x)$ is continuous
(c) Can we determine if $g(x)$ is twice-differentiable?
My solution:
(a) f defined Let $x\in\mathbb{R}$: then $|\frac{1}{k}\sin(\frac{x}{k})|\leq\frac{1}{k}|\sin(\frac{x}{k})|\leq\frac{1}{k}\frac{|x|}{k}=\frac{|x|}{k^2}$ (where we have used the fact that $|\sin(x)|\leq |x|$ for all $x\in\mathbb{R}$) and since $\sum_{k=1}^{\infty}\frac{|x|}{k^2}=|x|\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ by Comparison Test it must be $\sum_{k=1}^{\infty}|\frac{1}{k}\sin(\frac{x}{k})|<\infty$ which implies $\sum_{k=1}^{\infty}\frac{1}{k}\sin(\frac{x}{k})<\infty$ by Absolute Convergence Test. $f$ is thus well-defined for all $x\in\mathbb{R}$.
f continuous and differentiable $|f'_k(x)|=|\frac{1}{k^2}\cos(\frac{x}{k})|\leq\frac{1}{k^2}$ and $\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ so $\sum_{k=1}^{\infty} f'_k(x)$ converges uniformly on $\mathbb{R}$ by Weierstrass M-test and since $\sum_{n=1}^{\infty}f_n(0)=0$ by Term-by-Term Differentiability Theorem we have that $f$ si differentiable (and hence continuous) on $\mathbb{R}$ and $f'(x)=\sum_{k=1}^{\infty}f'_k(x)=\sum_{k=1}^{\infty}\frac{1}{k^2}\cos(\frac{x}{k})$.
f twice-differentiable $|f''_k(x)|=|-\sin(\frac{x}{k})\frac{1}{k^3}|\leq\frac{1}{k^3}$ and $\sum_{k=1}^{\infty}\frac{1}{k^3}<\infty$ so $\sum_{k=1}^{\infty}f''_k(x)$ converges uniformly on $\mathbb{R}$ by Weierstrass M-Test and since $\sum_{k=1}^{\infty} f'_k(x)=\sum_{k=1}^{\infty}\frac{1}{k}\cos(\frac{0}{k})=\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ by Term-by-Term Differentiability Theorem we have that $f'(x)$ is differentiable on $\mathbb{R}$  and $f''(x)=\sum_{k=1}^{\infty}f''_k(x)=\sum_{k=1}^{\infty}-\frac{1}{k^3}\sin(\frac{x}{k})$.
(b) g differentiable with g' continuous $|g'_k(x)|=|\frac{\cos(kx)}{k^2}|\leq\frac{1}{k^2}$ and $\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ so $\sum_{k=1}^{\infty}g'_k(x)$ converges uniformly on $\mathbb{R}$ by Weierstrass M-Test and since $\sum_{k=1}^{\infty}\frac{\sin(k\cdot 0)}{k^3}=0$ by Term-by-Term Differentiability Theorem $\sum_{k=1}^{\infty}g_k(x)$ converges uniformly to a differentiable function $g(x)=\sum_{k=1}^{\infty}g_k(x)=\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^3}$ on $\mathbb{R}$ and $g'(x)=\sum_{k=1}^{\infty}g'_k(x)=\frac{\cos(kx)}{k^2}$ and since each $g'_k$ is continuous by Term-by-Term Continuity Theorem $g'$ is continuous too.
g twice-differentiable? Here I have tried the estimates $|g''_k(x)|=|-\frac{\sin(kx)}{k}|\leq\frac{1}{k}$ which tells me nothing and neither does the estimate  $|g''_k(x)|=|-\frac{\sin(kx)}{k}|\leq\frac{k|x|}{k}=|x|$ since in both of these cases I can't use Weierstrass M-Test so I don't see how I could prove that $g$ is twice differentiable.

You need to use a partial summation/Dirichlet test to show that $h(x)=\sum_k \frac{\sin(kx)}{k}$ converges (locally uniformly, to a continuous function) away from $2\pi \Bbb{Z}$ (ie. using that $\sum_{k< K} \sin(kx)= \Im(\frac{e^{i Kx}-1}{e^{ix}-1})$ is bounded away from $2\pi \Bbb{Z}$)
Integrating $h$ two times (integrating termwise, allowed since the convergence is locally uniform) gives that $g$ is $C^2(\Bbb{R-2\pi Z})$.
