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.
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.
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 $\lim_{n\to \infty} u_n=0$
