Can I avoid the Abel partial summation technique and instead prove uniform convergence in this way? Edit:  I am attempting to use summation by parts...and will revise the below proof, as the R.H.S. says nothing more than what the L.H.S. already gives.
The problem statement is: 
For 2 sequences of functions ${a_n(x)}$, ${b_n(x)}$ on an interval $I$, suppose that $b_n(x)→0$ uniformly in $I$, $∑_1^{\infty}|b_n(x)−b_{n−1}(x)$| converges uniformly in $I$, and $∣∑_0^na_k(x)∣<M$ in $I$ for all $n$. Show that $∑a_n(x)b_n(x)$ converges uniformly in $I$.
My proof, using summation by parts:
To be less wordy, I won't show my derivation of the summation by parts formula, although it is the same spirit as robjohn's derivation on a link in MSE.
Since $b_n$ $\to 0$ uniformly On $I$, then for every $\epsilon >0$, there exists $N_1$ s.t. $n>N_1$ implies that $|b_n(x)|<\epsilon$.
Also, since $\sum |b_{n+1}(x) - b_n(x)|$ converges uniformly on $I$, then for every $\epsilon >0$, there exists $N_2$ such that $\sum_{N_2+1}^{\infty} |b_{n+1}(x) - b_n(x)|<\epsilon$, for all $x$ $\in$ $I$, by the Cauchy criterion for convergence.
Now, let $N=max(N_1, N_2)$.
We have that, by summation by parts, 
$$\sum_{n=N+1}^{m}a_n(x)bn(x) = (A_mb_m-A_{N}b_{N+1})-\sum_{n=N+1}^{m}An[b_{n+1}-b_n]$$
$$\implies |\sum_{n=N+1}^{m}a_n(x)bn(x)\large| = |(A_mb_m-A_{N}b_{N+1})-\sum_{n=N+1}^{m}An[b_{n+1}-b_n]|$$
$$\le |(A_mb_m| +|A_{N}b_{N+1}|+|\sum_{n=N+1}^{m}An[b_{n+1}-b_n]|$$
$$=M\epsilon + M\epsilon + M\epsilon$$
$$=3M\epsilon$$
since the partial sums of $a_n$ are bounded, i.e. $|A_n|<M$ for all $n$, using triangle inequality, and finally using the Cauchy criterion for the uniform convergence of the telescoping series of $b_n$.
Now, taking $m \to \infty$, the first estimate on the R.H.S. goes to zero, and we are left with the estimate 
$$|\sum_{n=N+1}^{\infty}a_n(x)bn(x)\large| \le M\epsilon + \sum_{n=N+1}^{\infty}An[b_{n+1}-b_n]|$$
$$\le M\epsilon +  M\epsilon$$
$$=2M\epsilon$$
But the inequality must hold for all $\epsilon>0$, and so we conclude that $\sum a_n(x)b_n(x)$ satisfies the Cauchy criterion for convergence, uniformly on $I$, and so the series must converge uniformly on $I$, since the space $R$ is a complete metric space.
End of proof.
My (erroneous) work, attempting to use the dominated convergence theorem:
We start with the series
$$\sum_{n=0}^{\infty}a_n(x)[b_n(x)-b_{n-1}(x)]= \lim_{m \to \infty}\sum_{n=0}^m a_n(x)\int_0^1(b_n(x)-b_{n-1}(x))dt$$
$$=\lim_{m \to \infty}\sum_{n=0}^m \int_0^1a_n(x)(b_n(x)-b_{n-1}(x))dt$$
$$=\lim_{m \to \infty}\int_0^1\sum_{n=0}^m a_n(x)(b_n(x)-b_{n-1}(x))dt$$
$$=\int_0^1\lim_{m \to \infty}\sum_{n=0}^m a_n(x)(b_n(x)-b_{n-1}(x))dt$$
$$=\int_0^1\sum_{n=0}^{\infty} a_n(x)(b_n(x)-b_{n-1}(x))dt$$
note that I took the limit inside of the integration, since here both the uniform convergence theorem or dominated convergence theorem would allow the switch.
Now I want to claim that the integrand is convergent, by the assumptions of uniform convergence of the telescoping series of $b_n$, coupled with the fact that the $a_n$ are bounded for all $n$.  Then the R.H.S. is finite.
And so the L.H.S. is finite and breaking out the L.H.S. gives two convergent series:
$$\sum_{n=0}^{\infty}a_n(x)[b_n(x)-b_{n-1}(x)] = \sum_{n=0}^{\infty}a_n(x)b_n(x) -\sum_{n=0}^{\infty}a_n(x)b_{n-1}(x)<\infty $$
And so we have that 
$$\sum_{n=0}^{\infty}a_n(x)b_n(x)<\infty$$
as required.  
Is this ok? 
Any hints or comments are welcome.
Thanks
 A: We use Abel partial summation to get the estimate of Cauchy sum of $\sum_{n=1}^{\infty} a_n(x)b_n(x)$.
Let $A_n(x)=\sum_{k=m}^n a_k(x)$. So for all $n,m$ and any $x\in I$, there is
$$
|A_n(x)|=\left|\sum_{k=1}^n a_k(x)-\sum_{k=1}^{m-1} a_k(x)\right|<2M\tag1
$$
We have
\begin{align}
\sum_{k=m}^n a_k(x)b_k(x)&=\sum_{k=m}^n (A_k(x)-A_{k-1}(x))b_k(x)
\\
&=\sum_{k=m}^n A_k(x)b_k(x) -\sum_{k=m}^n A_{k-1}(x)b_k(x)
\\
&=\sum_{k=m}^{n-1} A_k(x)(b_k(x)-b_{k+1}(x))+A_n(x)b_n(x)\tag{2}
\end{align}
Note that $A_{m−1}(x)=0$. 
By Cauchy criterian, for any $\epsilon>0$, there is a $N$ such that for $n,m>N$ and any $x\in I$, there is
$$
\sum_{k=m}^{n-1} |b_k(x)-b_{k+1}(x)|<\epsilon\quad\text{and }\quad |b_n(x)|<\epsilon\tag3
$$
So for all $n,m>N$ and any $x\in I$, by $(1), (2), (3)$ there is
\begin{align}
\left|\sum_{k=m}^n a_k(x)b_k(x)\right|&\leqslant\sum_{k=m}^{n-1} |A_k(x)||b_k(x)-b_{k+1}(x)|+|A_n(x)b_n(x)|
\\
&<2M\sum_{k=m}^{n-1} |b_k(x)-b_{k+1}(x)|+2M\:\epsilon 
\\
&<4M\epsilon
\end{align}
So by Cauchy Criterion, $\sum_{k=1}^{\infty} a_k(x)b_k(x)$ converges uniformly in $I$. 
