On the modes of convergence of the series $\sum\limits_{k=1}^{\infty} \frac{1}{k} \sin \left(\frac{x}{k+1} \right)$ Show that

$$f(x) = \sum_{k=1}^{\infty} \dfrac{1}{k} \sin \left(\dfrac{x}{k+1}\right).$$ 

converges pointwise on $\mathbb{R}$ and uniformly on each bounded interval in $\mathbb{R}$, to a differentiable function $f$ which satisifies $\vert f(x) \vert \leq \vert x \vert$ and $\vert f'(x) \vert \leq 1$.
This problem comes from a section that covers the Weierstrass M-Test and Dirichlet's Test. I am not sure where to begin with this. Analysis really isn't my thing at this point.

I'm sorry you're just getting your answer in five years. Lets have a look at this;
Let $[a,b]\subseteq \Bbb{R}$ be any bounded interval, $x\in [a,b]$ and $k\in \Bbb{N},$ then
\begin{align} \left|\dfrac{1}{k}\sin\left(\dfrac{x}{k+1}\right)\right|\leq  \dfrac{1}{k}\left|\dfrac{x}{k+1}\right|=\dfrac{\left|x\right|}{k(k+1)} \leq \dfrac{\left|x\right|}{k(k+1)} \leq \dfrac{c}{k(k+1)} ,  \end{align}
where $c:=\max\{|a|,|b|\}\quad$   [see If $x\in[a,b],$ then $|x|\leq \max\{|a|,|b|\}$. Since 
\begin{align} \sum^{\infty}_{k=1}\dfrac{c}{k(k+1)} ,  \end{align}
converges, then,
 \begin{align} \sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k+1}\right)   \end{align}
converges uniformly on $[a,b],$ by Weierstrass-M test. Since $[a,b]\subseteq \Bbb{R}$ was arbitrary, then uniform convergence is true on each bounded interval in $\Bbb{R}$ which also implies pointwise convergence on $\Bbb{R}$. Now, for arbitrary $x\in\Bbb{R},$
\begin{align} \left|f(x)\right|&=\left|\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k+1}\right)\right|\\&\leq \sum^{\infty}_{k=1} \dfrac{1}{k}\left|\dfrac{x}{k+1}\right|\\&= \sum^{\infty}_{k=1}\dfrac{\left|x\right|}{k(k+1)}\\&= \left|x\right|\sum^{\infty}_{k=1}\dfrac{1}{k(k+1)}\\&= \left|x\right|\lim\limits_{n\to\infty}\sum^{n}_{k=1}\dfrac{1}{k(k+1)} \\&= \left|x\right|\lim\limits_{n\to\infty}\sum^{n}_{k=1}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\\&= \left|x\right|\lim\limits_{n\to\infty}\left(1-\dfrac{1}{n+1}\right) \\&= \left|x\right|.  \end{align}
Since the series converges uniformly on $[a,b],$ then we can differentiate term-term on $[a,b].$ Thus, for $x\in [a,b]$
\begin{align} f'(x)&=\dfrac{d}{dx}\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k+1}\right)\\&=\sum^{\infty}_{k=1}\dfrac{1}{k}\dfrac{d}{dx}\left[\sin\left(\dfrac{x}{k+1}\right)\right]\\&=\sum^{\infty}_{k=1}\dfrac{1}{k}\left(\dfrac{1}{k+1}\right)\cos\left(\dfrac{x}{k+1}\right) ,  \end{align}
which implies that 
\begin{align} \left|f'(x)\right|&=\left|\sum^{\infty}_{k=1}\dfrac{1}{k}\left(\dfrac{1}{k+1}\right)\cos\left(\dfrac{x}{k+1}\right)\right|\\&\leq\sum^{\infty}_{k=1}\dfrac{1}{k}\left(\dfrac{1}{k+1}\right)\left|\cos\left(\dfrac{x}{k+1}\right) \right|\\&\leq\sum^{\infty}_{k=1}\dfrac{1}{k}\left(\dfrac{1}{k+1}\right)\\&= \lim\limits_{n\to\infty}\sum^{n}_{k=1}\dfrac{1}{k(k+1)} \\&= \lim\limits_{n\to\infty}\sum^{n}_{k=1}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\\&= \lim\limits_{n\to\infty}\left(1-\dfrac{1}{n+1}\right) \\&= 1   \end{align}
as required.
