Convergence of a sequence of convolutions 
Let $(a_n)$ be a sequence of real numbers such that
  $$
a_0>a_1>\cdots>0
$$
  and $M:=\sum_{n=0}^\infty a_n<+\infty$. Denote 
  $$
g_n=\frac{1}{a_n}\cdot 1_{[0,a_n]}
$$
   and define
  $$
f_n=g_0*g_1*\cdots*g_n
$$
  where $*$ means the convolutions. Show that $\{f_n\}$ converges pointwise. 


What I have done so far is that
$$
f_{n+1}(x)=f_n*g_{n+1}=\frac{1}{a_{n+1}}\int_{[x-a_{n+1},a_{n+1}]}f_n(y)\ dy. 
$$
and $f_n(x)=0$ for $x\not\in[0,M]$ and for all $n$. I don't see how to find a pointwise limit in $[0,M]$.
 A: Consider that every $g_n$ is the PDF of a random variable, say $X_n$. Then $f_n$ is the PDF of $X_1+\ldots+X_n$, and since $\text{supp}(f * g)=\text{supp}(f)+\text{supp}(g)$ and $\sum a_n$ is bounded, $\text{supp}(f_n)\subseteq [0,K]$ and $f_n$ is a compact-supported non-negative function with unit integral. Moreover, $f_n$ is the PDF of a unimodal, symmetric distribution and $f_n\in C^{(n-1)}(0,K)$.
However, a simple argument for proving the pointwise convergence without switching to Fourier transforms still eludes me, but I will keep thinking about it.
A simple way may be to provide some $\ell^1$ or $\ell^2$ bound for $\left|f_{n+1}(x)-f_n(x)\right|$ for every $x\in(0,K)$, as David Ullrich did.
A: One way or another show that $f_1$ is continuous. Now the Fundamental Theorem of calculus shows that $f_n$ is continuously differentiable for $n\ge2$, and in fact
$$f_n'(x)=\frac1{a_n}(f_{n-1}(x-a_n)-f_{n-1}(x))\quad(n\ge2).$$Let $$c_n=\sup_{x\in\Bbb R}|f_n'(x)|.$$The Mean Value Theorem shows that $c_n\le c_{n-1}$ for $n\ge3$, so we have $$|f_n'(x)|\le c\quad(n\ge 3).$$
I'm going to stop writing $n\ge2$ over and over. Now $$f_n(x)-f_{n-1}(x)=\frac1{a_n}\int_0^{a_n}(f_{n-1}(x-t)-f_{n-1}(x))\,dt,$$so again MVT shows that $$|f_n(x)-f_{n-1}(x)|\le\frac1{a_n}\int_0^{a_n}ct\,dt=\frac c2a_n.$$Since $\sum a_n<\infty$ this shows $f_n$ converges uniformly.

BONUS: In fact the limit is $C^\infty$. First recall a bit of calculus:
Lemma If $f_n\to f$ uniformly and $f_n'\to h$ uniformly then $f$ is differentiable and $f'=h$.
Now, for $n>2$ or $n>3$ or whatever "differentiation under the integral" (or various other arguments) shows that $$f_{n+1}'=g_{n+1}*f_n'.$$ So the same argument as showed that $f_n$ is uniformly convergent shows that $f_n'$ is also uniformly convergent, and then the lemma shows that $f=\lim f_n$ is differentiable. Similarly for higher derivatives, either by induction or by noting that $f_{n+1}^{(k)}=g_{n+1}*f_n^{(k)}$ for large $n$.
