Given that $\sum_{n \geq 1} a_n/n$ converges, Prove that $\lim_{n \to \infty} 1/n \sum_{k=1}^{n} a_k = 0$ [duplicate]

Let $s_0=0$ and for $n\geq 1$, $$s_n=\sum_{k=1}^{n}\frac{a_k}k$$ Then if $k\geq 1$, $a_k=k(s_k-s_{k-1})$

We have $$\begin{split} \sum_{k=1}^{n}a_k &= \sum_{k=1}^{n} k(s_k-s_{k-1})\\ &= \sum_{k=1}^{n} ks_k- \sum_{k=1}^{n}ks_{k-1}\\ &=\sum_{k=1}^{n} ks_k-\sum_{k=0}^{n-1} (k+1)s_k\\ &= ns_n -\sum_{k=1}^{n-1} s_k \end{split}$$ In other words $$\frac 1 n \sum_{k=1}^{n}a_k = s_n - \frac 1 n \sum_{k=1}^{n-1} s_k$$ Since by assumption, $s_n$ converges to a limit (let's call it $s$), by Cesaro's summation theorem, so does $\frac 1 n \sum_{k=1}^{n-1} s_k$. It follows that $$\lim_{n\rightarrow+\infty}\frac 1 n \sum_{k=1}^{n}a_k =s-s=0$$

Let $s_n=\sum_{k=1}^{n} \frac {a_k} k$. Then $a_n=n(s_n-s_{n-1})$. Hence $\frac 1 n \sum_{k=1}^{n} a_k=\frac 1 n (-s_1-s_2-\cdots-s_{n-1}+ns_n)$ (after some simplification). Now use the fact that $s_n \to s$ implies $\frac 1 {n-1} (-s_1-s_2-\cdots-s_{n-1}) \to -s$ and $\frac n {n-1} \to 1$.