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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.

mjw
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xldd
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  • Isn't it telescoping?

    \begin{align} \textstyle{\sum_{n \geq 1}} (a_n - a_{n+1})c_n &= c_1(a_1 - a_2) + c_2(a_2 - a_3) + \cdots \ &= a_1c_1 + a_2(c_2 - c_1) + \cdots \ &= \textstyle{\sum_{n \geq 1}} a_nb_n \end{align}

     – Riley Feb 26 '20 at 01:42
  • @Riley The partial sums differs by a term $a_nc_n$, which do not know to be convergent. So we could not take limit as you. – xldd Feb 26 '20 at 02:01
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    Oh right, my mistake, I just noticed what I have done is essentially the same as the attempt. In that case, if $a_n$ is unbounded, then there is a subsequence $a_{n_k} > 2^k$, so take $b_n$ along this subsequence as $1/2^k$ with zeroes everywhere else. Then $\sum b_n$ is convergent but $\sum a_nb_n$ is not. – Riley Feb 26 '20 at 02:11

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