Show 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 $m → \infty$. 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 $A_{n,m} = \Sigma_{k =m}^n a_k$
$\Sigma_{k = n}^m a_kb_k = A_{n,m}b_n - \Sigma_{k = m}^{n-1} A_{k,m}(b_{k+1} -b_k)$
 A: Let $c_k = a_kb_k$, $C_k =\sum_{j\ge k}c_j$. We can observe that
$$\begin{align*}
\sum_{k\ge m}a_k &=\sum_{k\ge m}\frac{c_k}{b_k}\\
&=\sum_{k\ge m}\frac{C_k-C_{k+1}}{b_k}\\
&=\lim_{N\to\infty}\sum_{k= m}^N\frac{C_k-C_{k+1}}{b_k}\\
&=\lim_{N\to\infty}\left(\sum_{k= m}^N\frac{C_k}{b_k}-\sum_{k= m+1}^{N+1}\frac{C_k}{b_{k-1}}\right)\\
&=\lim_{N\to\infty}\left(\sum_{k= m+1}^NC_k\left(\frac1{b_k}-\frac1{b_{k-1}}\right)+\frac{C_m}{b_m}-\frac{C_{N+1}}{b_N}\right)\\
&=\sum_{k= m+1}^\infty C_k\left(\frac1{b_k}-\frac1{b_{k-1}}\right)+\frac{C_m}{b_m}
\end{align*}$$ since $C_k\xrightarrow{k\to\infty} 0$ and $0\le b_k\le b_{k+1}\xrightarrow{k\to\infty} \infty \ \ (k\ge k_0)$. Then, we find that for $m\ge k_0$,
$$\begin{align*}
\left|b_m\sum_{k\ge m}a_k\right| &=\left|b_m\sum_{k= m+1}^\infty C_k\left(\frac1{b_k}-\frac1{b_{k-1}}\right)\right|+\left|C_m\right|\\
&\le  b_m\cdot \sup_{k\ge m+1}|C_k|\cdot \sum_{k\ge m+1} \left(\frac1{b_{k-1}}-\frac1{b_{k}}\right) +|C_m|\\
&\le \sup_{k\ge m+1}|C_k|+|C_m| \xrightarrow{m\to\infty} 0
\end{align*}$$ as wanted.
A: At first I assume that $$ a_{k},b_{k}\ge 0$$
We have 2 convergence based on input data:
$$ \sum_{k=1}^\infty a_{k}=C_{1} ,  \sum_{k=1}^\infty a_{k}b_{k}=C_{2} $$
we also assume that 
$$ \lim_{k=\infty} b_{k}=\infty  \Rightarrow $$ so for $$ m \rightarrow  \infty : b_{m}<b_{m+1}<b_{m+2}<... $$ 
So by simplifying mentioned equations:
$$ \sum_{k=1}^{m-1} a_{k}=D_{1} \Rightarrow D_{1}+\sum_{k=m}^\infty a_{k}=C_{1}  $$
$$ \sum_{k=1}^{m-1} a_{k}b_{k}=D_{2} \Rightarrow D_{2}+\sum_{k=m}^\infty a_{k}b_{k}=C_{2} $$
And we know that $$ C_{1},C_{2},D_{1},D_{2}<|M|<\infty$$ 
In conclusion I could say that :
$$ C_{2}-D_{2} =\sum_{k=m}^\infty a_{k}b_{k}=a_{m}b_{m}+a_{m+1}b_{m+1}+a_{m+2}b_{m+2}+...>b_{m}(a_{m}+a_{m+1}+a_{m+2}+...)$$
$$\Rightarrow \infty>|M_{2}|>C_{2}-D_{2}>b_{m}\sum_{k=m}^\infty a_{k}=b_{m}(C_{1}-D_{1})$$
$$\Rightarrow C_{2}-D_{2}>b_{m}(C_{1}-D_{1})$$
By having 
$$|M_{1}|>C_{1}-D_{1} , \lim_{k=\infty} b_{k}=\infty $$
I can receive to these:
$$ C_{1}-D_{1}=C_{2}-D_{2}=0 \Rightarrow \lim_{m=\infty}b_{m}\sum_{k=m}^\infty a_{k}=0$$
$$ $$
