Uniformly bounded partial sums of $\sum f'_n$ imply uniform convergence $\sum f_n$? Suppose $f_n: [a,b] \to \mathbb{R}$ is a sequence of differentiable functions where the series $\sum_n f_n(x_0)$ converges at one point. By Theorem 7.17 in Rudin (Principles of Mathematical Analysis) I conclude the series $\sum_n f_n(x)$ converges  uniformly to some function when the series $\sum_n f’_n(x)$ converges uniformly on $[a,b]$.
What if I know that $\sum_{n} f_n(x)$ converges for each $x \in [a,b]$ but just that partial sums $\sum_{n=0}^N f’_n(x)$ are bounded uniformly (but maybe not converging).  Is it still true that $\sum_n f_n(x)$ converges uniformly?
If not can someone please show me a counterexample.
 A: Uniform convergence of $\sum f_n'$ is much stronger than what is needed to guarantee uniform convergence of $\sum f_n$.  Uniform boundedness of the partial sums is sufficient.
Suppose $\sum_{n=0}^ \infty f_n(x)$ converges pointwise and 
$$\tag{*} \left|\sum_{n=0}^m f'_n(x)\right| \leqslant B$$ 
for all $x \in (a,b)$ and all $m \in \mathbb{N}.$  
Take a partition $a = x_0 < x_1 < \ldots x_r = b$ where $x_k - x_{k-1}  < \epsilon/(2B)$.  We have convergence of $\sum f_n(x)$  at the $r+1$ endpoints, and, hence,   there exists a positive integer $N$ such that for all $p > m > N$ we have 
$$|g(x_k)| := \left|\sum_{n = m+1}^{p} f_n(x_k) \right| < \frac{\epsilon}{2} \quad (k = 0,1,\ldots,r)$$
For any $x \in (a,b)$ there is a closest point $x_k$ such that $|x-x_k| < \epsilon/(4B)$. By the reverse triangle inequality and MVT, we have for some $\xi$ between $x$ and $x_k$,
$$|g(x)| \leqslant |g(x_k)| + |g(x) - g(x_k)| = |g(x_k)| + |g'(\xi)||x- x_k| \leqslant \frac{\epsilon}{2} + \frac{\epsilon}{4b} |g'(\xi)|$$
The inequality (*) leads to $|g'(\xi)| = \left|\sum_{n = m+1}^{p} f'_n(\xi) \right| \leqslant 2B$ and it follows that for all $x \in (a,b)$,
$$|g(x)| = \left|\sum_{n=m+1}^{p}f_n(x)\right| < \epsilon$$
