Prove that $\exists \{c_n\}$ monotonically increasing to $\infty$ such that $\sum_{i=1}^\infty a_nc_n$ coverges.

Here is an alternative approach. There is a sequence of positive integers $n_1<n_2<n_3<\cdots$ such that $\sum_{n=n_k}^\infty a_n < 2^{-k}$. Define $c_n = k$ when $n_k\leq n<n_{k+1}$ (with $n_0=1$). Then $c_n\nearrow+\infty$ and

$$\sum_n a_nc_n=\sum_{k=1}^\infty\sum_{n=n_k}^{n_{k+1}-1}a_nc_n\leq\sum_{k=1}^\infty k2^{-k}<\infty.$$

Put $\displaystyle R_n=\sum_{k\geq n} a_k$. Then $R_n>0$, and $R_n$ decrease to $0$ as $n\to +\infty$. We show that $\displaystyle c_n=\frac{1}{\sqrt{R_n}}$ (obviously increasing to $+\infty$) do the job.

We have: $$a_n c_n=\frac{a_n}{\sqrt{R_n}}=\frac{R_n-R_{n+1}}{\sqrt{R_n}}\leq \int_{R_{n+1}}^{R_n}\frac{dx}{\sqrt{x}}$$

Hence for $N \geq 1$:

$$\sum_{n=1}^N a_n c_n \leq \int_{R_{N+1}}^{R_1}\frac{dx}{\sqrt{x}}\leq \int_0^{R_1}\frac{dx}{\sqrt{x}}<+\infty$$ Hence $\sum a_n c_n$ is convergent.