If there exsits a constant M such that $ \forall x\in [a,b], \forall n\geq 1,|S'_n(x)|\leq M, $ then $\sum_{n=1}^\infty a_n(x)$ converges uniformly. 
Let series of functions $\sum_{n=1}^\infty a_n(x)$ converges on $[a, b]$, its partial sum are $S_n(x)$. If there exsits a constant M such that
$$
\forall x\in [a,b], \forall n\geq 1,|S'_n(x)|\leq M,
$$
then $\sum_{n=1}^\infty a_n(x)$ converges uniformly.

I tried to prove the problem by Cauchy criterion.
$$
S_{n+p}(x)-S_n(x)=S_{n+p}'(\xi_1)(x-a)+S_{n+p}(a)-S_{n}'(\xi_2)(x-a)-S_{n}(a)
$$
But it seems we are unable to to control $|S'_{n+p}-S'_n|$, so I  amstuck here.
Is my way wrong, if so, please suggest a correct way. Appreciate any help!
 A: We have $S_n(x) \to S(x)$ pointwise and $|S_n(x) - S_n(y)| \leqslant |S_n'(\xi)||x-y| \leqslant M|x-y|$ for all $x,y \in [a,b]$ and all $n \in \mathbb{N}$.
For all $\epsilon > 0$ there exists $\delta > 0$ such that if $|x-y| < \delta$, then for all $n \in\mathbb{N}$ we have $|S_n(x) - S_n(y)| < \epsilon/3$. Taking the limit as $n \to \infty$ it follows that $|S(x) - S(y)| \leqslant \epsilon/3$ as well.
Since $[a,b]$ is compact there exist finitely many points $x_1, \ldots x_m$ such that $[a,b] \subset \cup_jI(x_j,\delta),$ where $I(x_j,\delta)$ is the open interval with center $x_j$ and radius $\delta$.
By pointwise convergence, there exists $N_j \in \mathbb{N}$ such that if $n \geqslant N_j$ then $|S_n(x_j) - S(x_j)|< \epsilon/3.$ Let $N = \max_{1 \leqslant j \leqslant m}N_j.$
Given any $x \in [a,b]$ there exists $j$ such that $x \in I(x_j,\delta)$. This implies $|x - x_j| < \delta$ and if $n \geqslant N$ we have
$$|S_n(x) - S(x)| \leqslant |S_n(x) - S_n(x_j)|+|S_n(x_j) - S(x_j)|+|S(x_j) - S(x)| \\\leqslant \frac{\epsilon}{3} +  \frac{\epsilon}{3}+  \frac{\epsilon}{3} = \epsilon,$$
proving uniform convergence.
