Q:

Find limit of sum

I suspect that $\lim_{n \to \infty} \sum_{k = 0}^{n  - 1}\frac{k}{a^k(n - k)} = 0$ for $a > 1$. I know that this product represents the taylor coefficients of $\frac{-ax\ln(1 - x)}{(a - x)^2}$ by the Cauchy product. Unfortunately the limit as $x \to 1^{-1}$ is not defined so I can't use Abel's theorem. How can I prove this?

A:

The limit is zero as expected. First, write the sum as
$$ \sum_{k=0}^{n-1} \frac{k}{a^k(n-k)} = \sum_{k=0}^{\infty} \frac{k}{a^k(n-k)} \mathbf{1}_{\{k < n\}}$$
Since each term is uniformly bounded by
$$ \left| \frac{k}{a^k(n-k)}\mathbf{1}_{\{k < n\}} \right| \leq \frac{k}{a^k} $$
and $\sum_{k=0}^{\infty} \frac{k}{a^k} < \infty$, the dominated convergence theorem tells that
$$ \lim_{n\to\infty} \sum_{k=0}^{\infty} \frac{k}{a^k(n-k)} \mathbf{1}_{\{k < n\}} 
= \sum_{k=0}^{\infty} \lim_{n\to\infty} \frac{k}{a^k(n-k)} \mathbf{1}_{\{k < n\}}
= \sum_{k=0}^{\infty} 0 = 0. $$

