For all $\sum b_n$ convergent, $\sum a_nb_n$ convergent. Show $\sum |a_n-a_{n+1}|$ convergent.
I have found If $\sum a_nb_n$ converges for each null sequence $(b_n)$ then $\sum a_n$ is absolutely convergent being of helpful.
For any $\lim c_n=0$, define $b_1=c_1, b_n=c_n-c_{n-1}, n\geq 2$, then $\sum b_n$ convergent. By the assumption, $\sum a_nb_n$ convergent.
But how to prove $\sum (a_n-a_{n-1})c_n$ convergent? So that If $\sum a_nb_n$ converges for each null sequence $(b_n)$ then $\sum a_n$ is absolutely convergent can be applied.
My attempt. By the Abel transform, $$\sum_{k=1}^n a_kb_k=a_nc_n-a_1c_0+\sum_{k=1}^{n-1}(a_k-a_{k-1})c_k.$$
How to know first $a_n$ is bounded, so that we can deduce $\sum (a_n-a_{n-1})c_n$ convergent.