TITLE: How to show uniform convergence of this series of functions
QUESTION [1 upvotes]: I was working the following question from a previous qualifying exam, the solution below seems to have a minor snag in it below

Let $f_{1}: [a,b] \rightarrow \mathbb{R}$ be Riemann integrable function. Define the sequence of functions $f_{n}: [a,b] \rightarrow \mathbb{R}$ by 
  $$ f_{n+1}(x)= \int^{x}_{a} f_{n}(t)dt $$ for each $n \geq 1$ and $x \in [a,b]$. Prove that the sequence of functions $$g_{n}(x)= \sum^{n}_{m=1} f_{m}(x)$$ converges uniformly on $[a,b]$. 

We seek to show that sequence of sums $(f_{1}, f_{1} +f_{2}, \ldots)$ converges uniformly. We observe that since $f_{1}$ is Riemann integrable, it is bounded meaning that if $x\in [a,b]$ then $|f_{1}(x)| \leq k$ for some postive real number $k$. It follows that $|f_{n}(x)| \leq k(x-a)^{n-1} \leq k(b-a)^{n-1}$. 
The Weierstrass $M$ test, then tells us that if $|f_{n}(x)| \leq M_{n}$ and $\sum M_{n}$ converges then the sequence $(f_{1}, f_{1} +f_{2}, \ldots)$ converges uniformly.  In this case, we have $M_{n}= k(b-a)^{n-1}$, but, I believe that $k \sum (b-a)^{n-1}$ converges only when $|b-a|<1$.  
Is what I have done so far correct, or is there another way to do this to move around this snag of the length of the interval.
Thanks!
Epilogue
Daniel Fischer in the comments helped me find a better estimate than the one before. We can actually find that $|f_{n}(x)| \leq k (b-a)^{n-1}/(n-1)!$. In which case the series $\sum M_{n}$ does converge with respect to the ratio test. To see this,
We can estimate 
$|f_{2}(x)| \leq \int^{x}_{a} |f_{1}(t)|dt \leq k(x-a)$
and then
$|f_{3}(x)| \leq \int^{x}_{a} |f_{2}(t)|dt \leq \int^{x}_{a} k(t-a) dt= k(x-a)^{2}/2$
and then 
$|f_{4}(x)| \leq \int^{x}_{a} |f_{3}(t)|dt \leq \int^{x}_{a} \frac{k}{2}(t-a)^{2} dt= k(x-a)^{3}/6$
and so forth....

REPLY [1 votes]: Your original work has been correct, but as you noticed, obtaining a global bound for $f_{n+1}$ from a global bound for $f_n$ via
$$\lvert f_{n+1}(x)\rvert = \biggl\lvert \int_a^x f_n(t)\,dt\biggr\rvert \leqslant \int_a^x \lVert f_n\rVert_{\infty}\,dt = \lVert f_n\rVert_{\infty}\cdot (x-a) \leqslant \lVert f_n\rVert_{\infty}\cdot (b-a)$$
doesn't yield a sharp enough bound for $b-a \geqslant 1$.
As you then found after the hint in the comments, using a pointwise bound on $f_n$ to obtain a pointwise bound for $f_{n+1}$ via
$$\lvert f_{n+1}(x)\rvert \leqslant \int_a^x \lvert f_n(t)\rvert\,dt \leqslant k\int_a^x \frac{(t-a)^{n-1}}{(n-1)!}\,dt = k\frac{(x-a)^n}{n!}$$
inductively yields the global bound $\lVert f_n\rVert_{\infty} \leqslant k\dfrac{(b-a)^n}{n!}$ which is sharp enough to yield the uniform convergence per the Weierstraß $M$-test on any interval $[a,b]$.
Since in the case of a constant $f_1$ all the inequalities are equalities, these bounds are optimal.
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