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

Question

I came in contact with the following problem:

Be a sequence $(a_n)_{n\geq1}$ a sequence of real numbers, given that $\sum_{n\geq1} a_n/n $ converges, prove that $\lim_{n \to \infty} 1/n \sum_{k=1}^{n} a_k = 0$.

Now, I want to say that I know how to do if $1/n$ was replaced by $1/n^2$, I would just replace $\sum_{k=1}^{n} a_k$ with $s_n$ and use the fact that $\sum_{n\geq1} a_n/n $ converges, this is, $\sum_{n\geq1} a_n/n \leq \epsilon$ for any $\epsilon > 0$, there exists $N$ such that this happens. And we would have:

$$s_n/n \leq S_N/n + \sum_{m>N}^{n} a_m/n \leq s_N/n + \sum_{m>N} a_m/m < s_n/n + \epsilon$$

Is it this way that I would solve?

Your notation choices are a bit confusing. When you write "$\sum_{n\geq 1}a_n / n\leq \epsilon$", is the $\sum_{n\geq 1}a_n / n$ meant to denote the $n$th partial sum of the series $\sum_{k=1}^\infty a_k / k$? If so, then I'd recommend editing your post to replace your notation with $\sum_{k=1}^n a_k / k$. — Alann Rosas, Jan 07 '22 at 09:55
This follows from a slightly more general result, known as Kronecker's lemma, with $x_k=\frac{a_k}{k}$ and $b_k=k$. — peek-a-boo, Jan 07 '22 at 09:57
Also, you aren't justified in writing "$\sum_{n\geq 1} a_n / n\leq\epsilon$", regardless if $\sum_{n\geq 1}a_n /n$ denotes the partial sum or the limit of the series. This suggests that said limit is zero or less, but you weren't given this information. All you know from the convergence of $\sum_{k=1}^\infty a_l / k$ is that the partial sums have some limit, say $L$, with the property that for every $\varepsilon>0$, there is a $N\in\mathbb{N}$ such that for every $n$ with $n>N$, $\left|\sum_{k=1}^n a_k /k -L\right|<\varepsilon$. — Alann Rosas, Jan 07 '22 at 10:02
There are some older posts about this problem on this site, such as: How to show that $\lim_{n\to \infty} \frac{a_1 +a_2 + \cdots + a_n}{n} = 0?$, If $\sum_{n\geq 1}\frac{a_n}{n}$ converges, then $\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^na_k=0$. I will also add this link, in case it might help sometime: How to search on this site? — Martin Sleziak, Jan 07 '22 at 10:59

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

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