Limit of convergent series equal zero 
Show that if $\sum_{n=1}^{\infty} a_n$ is convergent then
  $$
\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n k a_k=0
$$

If we let the limit of series be $A$ then it can be shown $\lim_{n\to\infty} (a_1+\dots + a_n)/n=A$ as well. But how to deal with the term $ka_k$?

Let $A_n=\sum_{k=1}^n a_k\to A$ then, by summation by parts,
$$\sum_{k=1}^{n} ka_k=\sum_{k=1}^{n} k(A_k-A_{k-1})=\sum_{k=1}^{n} kA_k-\sum_{k=1}^{n}(k-1)A_{k-1}-\sum_{k=1}^{n}A_{k-1}=nA_n-\sum_{k=1}^{n}A_{k-1}$$
Hence, by the Stolz-Cesaro theorem,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} ka_k=\lim_{n\to\infty}A_n
-\lim_{n\to\infty}\frac{\sum_{k=1}^{n}A_{k-1}}{n}=
A-\lim_{n\to\infty}\frac{A_n}{(n+1)-n}=A-A=0.$$
