$n\lvert a_n\rvert \to 0$, then $\sum_{k=1}^n \frac k n\lvert a_k\rvert \to 0$?

Question

I am now working on the converse of Abel's THM and found out one proof of the conditional converse of the theorem. The proof says,

Suppose $$n\lvert a_n\rvert \to 0$$ as $$n \to \infty$$

Then, $$\sum_{k=1}^n \frac k n\lvert a_k\rvert \to 0$$ as $$n \to \infty$$ But how does this work? Should I use comparison test? I tried to use root test but I found it hard to apply it given the form of the series. Can you help me? I'd really appreciate your help.

Answer 1

If $$ n\lvert a_n\rvert \to 0, $$ then $$ \frac 1n\sum_{k=1}^n k\lvert a_k\rvert \to 0, $$ This may be seen as an application of the Cesaro theorem.

Answer 2

Fix $\varepsilon>0$. Then there exist $m $ such that $k|a_k|<\varepsilon/2$ when $k>m $. There also exists $c $ with $k|a_k|\leq c $ for all $k $ (a convergent sequence is bounded). Now if $$n>\max\left\{m, \frac {2mc}\varepsilon\right\},$$you have \begin{align}\sum_{k=1}^n\frac kn \,|a_k| &= \sum_{k=1}^m\frac kn \,|a_k|+ \sum_{k=m+1}^n\frac kn \,|a_k|\leq\sum_{k=1}^m \frac {c}n+\frac\varepsilon2\,\sum_{k=m+1}^n\frac1n\\ \ \\ &=\frac {mc}n+\frac {\varepsilon\, (n-m)}{2n} \leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon. \end{align}