Suppose that $$\lim_{n\to\infty}\sum_{k=0}^nk|x_k|=L<\infty.$$ Is it possible to show that $$\lim_{n\to\infty}\sum_{k=0}^n|x_k|=K<\infty.$$ In words, does the absolute summability of $\{nx_n\}_{n=0}^\infty$ imply the absolute summability of $\{x_n\}_{n=0}^\infty$?
The doubt comes from time-series. For simplicity, say that we have a mean-zero strongly stationary process $\{X_t\}_t$. It is often the case that the auto-covariance function $\gamma: \mathbb{N}\to \mathbb{R}, k\mapsto \mathbb{E}[X_{t}X_{t-k}]$ is restricted to have a controlled decay as $k$ grows large. Two (of the many) ways of achieving that are $$\lim_{T\to\infty}\sum_{k=0}^T|\gamma(k)|<\infty\qquad\text{and}\qquad \lim_{T\to\infty}\sum_{k=0}^T|k\gamma(k)|<\infty, $$ where the fact that $\gamma(\cdot)$ is an even function automatically gives $$\lim_{T\to\infty}\sum_{k=-T}^T|\gamma(k)|<\infty\qquad\text{and}\qquad \lim_{T\to\infty}\sum_{k=-T}^T|k\gamma(k)|<\infty. $$
Typically the latter is deemed as stronger, but I have never seen a formal proof nor been able to do it myself.
This is straightforward. Every single summand satisfies $$ k|x_k| \geq |x_k| $$ (except for $k=0$, but the value of finitely many values does not have an impact on convergence of a series), so their partial sums satisfy $$ \sum_{k=0}^n k|x_k| = \sum_{k=1}^n k|x_k|\geq \sum_{k=1}^n |x_k| $$ for every $n>0$, so this also holds in the limit: $$ L-|x_0|=\lim_{n\to\infty}\sum_{k=1}^n k|x_k| \geq \lim_{n\to\infty}\sum_{k=1}^n |x_k|. $$ And since the right hand partial sums are monotonically increasing (since all summands are nonnegative), they must converge to some $K\leq L-|x_0|$.