Convergent series question - Dirichlet's Test? 
Proposition Suppose $\{a_n\}$ and $\{b_n\}$ are real sequences such that $a_n \to 0$.  Show that $\sum a_kb_k$ converges under the following two conditions: $\sum_{k=1}^{\infty} |a_{k+1}-a_k|< \infty$ and $|\sum_{k=1}^{n} b_k| \leq M$.

I'm stuck on this proof because I'm not sure how this: $\sum_{k=1}^{\infty} |a_{k+1}-a_k|< \infty$ fits into the bigger picture.  In terms of finite sums this would be a telescoping series.

Suppose that $\sum_k|a_{k+1}-a_k|<\infty$ and $|\sum_{k=1}^nb_k|\leq M$ (I will assume that this last condition happens for all $n$). 
Using Newton series, for $n>m$,
$$
\sum_{k=m}^na_kb_k=a_n\sum_{k=m}^nb_k-\sum_{k=m}^{n-1}(a_{k+1}-a_k)\sum_{h=m}^kb_h,
$$
so, 
$$
\left|\sum_{k=m}^na_kb_k\right|\leq\,|a_n|\,\left|\sum_{k=m}^nb_k\right|+\sum_{k=m}^{n-1}|a_{k+1}-a_k|\,\left|\sum_{h=m}^kb_h\right|\leq\,M\,|a_n|+M\,\sum_{k=m}^{n-1}|a_{k+1}-a_k|.
$$
Given $\varepsilon>0$, we can choose $m$ such that $|a_n|<\varepsilon/2M$, $\sum_{k=m}^n|a_{k+1}-a_k|<\varepsilon/2M$, which yields
$$
\left|\sum_{k=m}^na_kb_k\right|<\varepsilon.
$$
So the series converges. 
