Edit: I revised the choice of $N=max(N_1,N_2)$, in order to give slightly sharper estimates to make sure I can say "less than epsilon" at the end, and not be stuck with a 3M$\epsilon$.
Is my proof ok? In particular, was it necessary to let $m \to \infty$ and give a second estimate? Thanks,
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 $I$.
Proof, using summation by parts:
Since $b_n$ $\to 0$ uniformly On $I$, then for every $\epsilon >0$, there exists $N_1$ s.t. $n>N_1$ implies that $|b_n(x)|<\large \frac{\epsilon}{3M}$.
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 $\sum_{N_2+1}^{\infty} |b_{n+1}(x) - b_n(x)|<\large \frac{\epsilon}{3M}$, 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,
$$\sum_{n=N+1}^{m}a_n(x)bn(x) = (A_mb_m-A_{N}b_{N+1})-\sum_{n=N+1}^{m}An[b_{n+1}-b_n]$$
$$\implies |\sum_{n=N+1}^{m}a_n(x)bn(x)\large| = |(A_mb_m-A_{N}b_{N+1})-\sum_{n=N+1}^{m}An[b_{n+1}-b_n]|$$
$$\le |(A_mb_m| +|A_{N}b_{N+1}|+|\sum_{n=N+1}^{m}An[b_{n+1}-b_n]|$$
$$=M\epsilon/3M + M\epsilon/3M + M\epsilon/3M$$ $$=\epsilon$$
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
$$|\sum_{n=N+1}^{\infty}a_n(x)bn(x)\large| \le M\epsilon/3M + \sum_{n=N+1}^{\infty}An[b_{n+1}-b_n]|$$
$$\le M\epsilon/3M + M\epsilon/3M$$
$$<\epsilon$$
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.