real analysis - Show that if $b_kuparrow infty$ and $Sigma_{k = 1}^infty a_kb_k$ converges, then $b_m Sigma_{k = m}^infty a_k → 0$ as $m → infty$.

Suppose that $\Sigma_{k=1}^\infty a_k$ converges. Prove that if $b_k\uparrow \infty$ and $\Sigma_{k = 1}^\infty a_kb_k$ converges, then

$b_m \Sigma_{k = m}^\infty a_k → 0$ as $m → \infty$.

Attemtp: Suppose $\Sigma_{k=1}^\infty a_k$ converges, and $\Sigma_{k = 1}^\infty a_kb_k$ also.Then we know by Abel's Formula that the sequences converge only iff its partial sums converge. If we let $\Sigma a_k = \Sigma \frac{a_kb_k}{b_k}$, then because $b_k$ is increasing we have $\frac{1}{b_k}$ is decreasing to zero. So we almost have a telescoping series.

Then let $c_k = \Sigma_{ j = k}^{\infty} a_jb_j$. Then $\Sigma_{k = n}^m a_k = \Sigma \frac{a_kb_k}{b_k}= \Sigma_{k = n}^m \frac{c_k - c_{k+1}}{b_k}$

I don't know how to continue. I am having trouble applying Abel's Formula and the subscripts are confusing. Can someone please help me? I am suppose to use Abel's Formula. Thank you very much.

Abel's Formula: Let $a_k,b_k$ be real sequences, and for each pair of integers $n \geq m \geq 1$ set $A_{n,m} = \Sigma_{k =m}^n a_k$
$\Sigma_{k = n}^m a_kb_k = A_{n,m}b_n - \Sigma_{k = m}^{n-1} A_{k,m}(b_{k+1} -b_k)$