Show that $f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin(\frac{x}{k+1})$ converges. Exercise: Show that
$$f(x)=\sum_{k=1}^\infty \frac{1}{k}\sin\left(\frac{x}{k+1}\right)$$
converges, pointwise on $\mathbb{R}$ and uniformly on each bounded interval in $\mathbb{R}$, to a differentiable function $f$ which satisfies
$$|f(x)|\leq |x| \text{ and } |f'(x)|\leq 1$$
for all $x\in \mathbb{R}$.
Hint: Dominate, then telescope.
I am unsure how to start this proof. By definition, I know that I need to show that the sequence with the terms
$$s_n(x)=\sum_{k=1}^n \frac{1}{k}\sin\left(\frac{x}{k+1}\right)$$
converges pointwise on $\mathbb{R}$ and uniformly on a bounded interval in $\mathbb{R}$, but I am unsure how to show these facts.
 A: Let's denote by
$$f_n(x)=\frac{1}{n}\sin(\frac{x}{n+1}),$$
so we have $$f(x)=\sum_{n=1}^\infty f_n(x).$$
It's clear that $f$ is defined at $0$ and
$$f_n(x)\sim\frac{x}{n^2},\forall x\neq0$$
then we have pointwise convergence by comparaison with the Riemann series.
Now, let $[a,b]$ a bounded interval in $\mathbb{R}$. We have
$$|f_n(x)|\leq|\frac{x}{n^2}|\leq \frac{\max(|a|,|b|)}{n^2}, $$
so we have normal convergence which implies the uniform convergence of the series on $[a,b]$.
Moreover, from
$$|f'_n(x)|=\left|\frac{\cos(\frac{x}{n+1})}{n(n+1)}\right|\leq\frac{1}{n^2},$$
hence, we find the uniform convergence of the series $\sum_nf'_n(x)$ on $\mathbb{R}$ which prove that $f$ is differentiable and
$$f'(x)=\sum_{n=1}^\infty f'_n(x).$$
Finally, we have these two inequality
$$|f(x)|\leq\sum_{n=1}^\infty \frac{1}{n}|\sin\frac{x}{n+1}|\leq|x|\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})=|x|,$$
and
$$|f'(x)|\leq\sum_{n=1}^\infty|f'_n(x)|\leq\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})=1.$$
A: Compare with $$\sum_{k=1}^\infty \dfrac{x}{k(k+1)}.$$
