Suppose $f'$: (-1,1) → R and is bounded where $f(0) = 0$. Show that $$\sum_{n=1}^\infty \frac{1}{n}f\left(\frac{x}{n+1}\right)$$ converges uniformly on (-1,1) and $S'(0)=f'(0)$.
I believe I have to use Dirichlet's Test for Uniform Convergence (because $\frac{1}{n}$ is decreasing to $0$ and converges uniformly), but I do not know how to show $\sum f\left(\frac{x}{n+1}\right)$ is bounded.
Edit: using the link; I used the Weierstrass M-test to show uniform convergence as it made the most sense, but now I am struggling with showing $S'(0)=f'(0)$. I would assume you would do term by term differentiation to get $$ S'(x) = \sum_{n=1}^\infty \frac{1}{n(n+1)}f'\left(\frac{x}{n+1}\right)$$ and then it would suffice to show that $\sum \frac{1}{n(n+1)} = 1$, which is where I get stuck.
hint
For $x\in(-1,1)$, use MVT to get
$$f(\frac{x}{n+1})-f(0)=\frac{x}{n+1}f'(C_x)$$
and
$$|f(\frac{x}{n+1})|<\frac{M}{n}$$