In case the sequence $|a_k|$ is non-decreasing then one can show that $ka_k\to 0$ and the result would follow but what if we dont have the monotonicity?
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Let $A_n=a_0+a_1+\ldots+a_n$. By summation by parts:
$$ \frac{1}{n}\sum_{k=0}^{n}k a_k = A_n - \frac{1}{n}\sum_{k=0}^{n-1} A_k $$ and we know that $\lim_{n\to +\infty}A_n = L<+\infty$, since the absolute convergence of $\sum_{n\geq 0}a_n$ implies the conditional convergence. By Cesàro theorem $\lim_{n\to +\infty}A_n=L$ implies $\lim_{n\to +\infty}\frac{A_0+A_1+\ldots+A_{n-1}}{n}=L$, hence $$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=0}^{n}k a_k = L-L = 0 $$ as wanted.
Jack D'Aurizio
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Given $\varepsilon > 0$, there exists $N\in\mathbb{N}$ such that $\sum_{k=N+1}^\infty |a_k| < \varepsilon$.
If $n> N$ you have that $$ \frac{1}{n} \sum_{k=0}^n k |a_k| \leq \frac{1}{n}\sum_{k=0}^N k |a_k| + \sum_{k=N+1}^n |a_k| < \frac{1}{n}\sum_{k=0}^N k|a_k| + \varepsilon. $$ For $n\to +\infty$, the first summation at r.h.s. goes to $0$, hence $$ \limsup_{n\to +\infty} \frac{1}{n} \sum_{k=0}^n k |a_k| \leq \varepsilon. $$
Rigel
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