real analysis - How to prove this sequence is convergent?

Suppose that the series $\sum_{k=1}^\infty{a_k}$ converges. Prove that
$$\lim_{n→\infty}\frac{1}{n}\sum_{k=1}^{n}ka_k=0$$

I tried to use the definition of convergence of $\sum a_k$ but I'm struggling to find $N$ such that given $\epsilon >0$, if $n>N$, then $-\epsilon<\frac{1}{n}\sum_{k=1}^{n}ka_k<\epsilon$ but I don't know how to connect them

hope somebody can help

Answer

Let $s_n=a_1+\cdots+a_n\to a$. Then
$$
\sum_{k=1}^n ka_k=\sum_{k=1}^n k(s_k-s_{k-1})=\sum_{k=1}^n ks_k-\sum_{k=1}^{n-1}(k+1)s_k=ns_n-\sum_{k=1}^{n-1}s_k.
$$
Hence
$$
\frac{1}{n}\sum_{k=1}^n ka_k=s_n-\frac{n-1}{n}\cdot\frac{1}{n-1}\sum_{k=1}^{n-1}s_k\to a-a=0.
$$
We have used the fact that: If $b_n\to b$, then so does $\,\,\dfrac{b_1+\cdots+b_n}{n}$.

Note. However, it is not in general true that $na_n\to 0$.