How to show that $\lim_{n\rightarrow \infty}a_n^{-1}\sum_{k=1}^{n}a_kc_k=0$. I'm trying to solve the following problem: 

If $\left(\, a_{n}\,\right)$ is a positive monotonically increasing and diverging sequence of real numbers and $\left(\, c_{n}\,\right)$
  is a summable sequence, then
  $\lim_{\, n \to \infty}\,
\left(\, a_{n}^{-1}\sum_{k = 1}^{n}a_{k}\, c_{k}\,\right) = 0$.

I can't figure out how to proceed or to use the given assumptions on divergence and summability. I'd appreciate some help.
 A: If "summable sequence" is to be understood in the sense of general summability (see also), then $\sum_{k = 1}^\infty \lvert c_k\rvert < +\infty$, and we have a short proof by splitting at a fixed index $m$:
$$\Biggl\lvert\, a_n^{-1}\sum_{k = 1}^n a_kc_k\Biggr\rvert \leqslant a_n^{-1} \sum_{k = 1}^m a_k\lvert c_k\rvert + \sum_{k = m+1}^n \frac{a_k}{a_n}\lvert c_k\rvert \leqslant a_n^{-1}\sum_{k = 1}^m a_k\lvert c_k\rvert + \sum_{k = m+1}^n \lvert c_k\rvert,$$
and since $a_n \to +\infty$, we have
$$\lim_{n\to\infty} a_n^{-1}\sum_{k = 1}^m a_k\lvert c_k\rvert = 0,$$
from which we deduce
$$\limsup_{n\to\infty}\; \Biggl\lvert\, a_n^{-1}\sum_{k = 1}^n a_k c_k\Biggr\rvert \leqslant \sum_{k = m+1}^\infty \lvert c_k\rvert.$$
That holds for all $m$, and therefore
$$\limsup_{n\to\infty}\; \Biggl\lvert\, a_n^{-1}\sum_{k = 1}^n a_k c_k\Biggr\rvert \leqslant \lim_{m\to\infty} \sum_{k = m+1}^\infty \lvert c_k\rvert = 0.$$
If "summable sequence" is to be understood merely as the existence of
$$\lim_{n\to\infty} \sum_{k = 1}^n c_k,$$
i.e. possibly conditional convergence of the series $\sum c_k$, then we must argue more carefully. Since $(a_n)$ is monotonic, summation by parts leads to the goal. For $n \geqslant 1$, we define
$$r_n := \sum_{k = n}^\infty c_k\qquad \text{and}\qquad s_n := \sup \{ \lvert r_k\rvert : k \geqslant n\}.$$
The (conditional) convergence of $\sum_{k = 1}^\infty c_k$ ensures that $r_n$ is well-defined, and $r_n \to 0$, which also entails $s_n \to 0$. Then for an arbitrary $m \in \mathbb{N}$ and $n > m$, we have
\begin{align}
\sum_{k = m+1}^n a_k c_k &= \sum_{k = m+1}^n a_k(r_k - r_{k+1}) \\
&= \sum_{k = m+1}^n a_k r_k - \sum_{k = m+2}^{n+1} a_{k-1} r_k \\
&= a_{m+1} r_{m+1} - a_n r_{n+1} + \sum_{k = m+2}^n (a_k - a_{k-1})r_k,
\end{align}
and therefore
\begin{align}
\Biggl\lvert \sum_{k = m+1}^n a_k c_k\Biggr\rvert &\leqslant a_{m+1}\lvert r_{m+1}\rvert + a_n\lvert r_{n+1}\rvert + \sum_{k = m+2}^n (a_k - a_{k-1})\lvert r_k\rvert \\
&\leqslant a_{m+1}s_{m+1} + a_ns_{n+1} + \sum_{k = m+2}^n (a_k - a_{k-1})s_k \\
&\leqslant s_{m+1}\cdot\Biggl(a_{m+1} + a_n + \sum_{k = m+2}^n (a_k - a_{k-1})\Biggr)\\
&= 2s_{m+1}a_n.
\end{align}
Then we estimate
$$\Biggl\lvert\, a_n^{-1}\sum_{k = 1}^n a_kc_k\Biggr\rvert \leqslant a_n^{-1} \sum_{k = 1}^m a_k\lvert c_k\rvert + a_n^{-1}\Biggl\lvert\sum_{k = m+1}^n a_k c_k\Biggr\rvert \leqslant a_n^{-1} \sum_{k = 1}^m a_k\lvert c_k\rvert + 2s_{m+1}$$
and conclude
$$\limsup_{n\to\infty}\; \Biggl\lvert\, a_n^{-1}\sum_{k = 1}^n a_k c_k\Biggr\rvert = 0$$
as above.
A: Since $\sum c_n<\infty \implies \lim c_n=0$ 
Now the  $n^{th}$ term of the given sequence 
$u_n=\dfrac{a_1c_1+a_2c_2+\ldots +a_nc_n}{a_n}=(\dfrac{a_1}{a_n})c_1+(\dfrac{a_2}{a_n})c_2+\ldots +(\dfrac{a_n}{a_n})c_n=\sum _{k=1}^nb_kc_k$
where $b_k=(\dfrac{a_k}{a_n})<1$
So $\lim_{n\to \infty}u_n=\lim_{n\to \infty}\sum _{k=1}^nb_kc_k=\sum_{k=1}^\infty b_kc_k$
Since $a_n$ is unbounded so for each $n\in \Bbb N\exists a_n$ such that $a_n\ge n^2\implies \dfrac{1}{a_n}\le \dfrac{1}{n^2}$
Also $\lim c_n=0\implies \exists p\in \Bbb N$ such that $|c_n|\le \frac{1}{n}\forall n\in \Bbb N$
Combining two we have $\lim_{n\to \infty} u_n=0$
