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 $ka_k$?

We want to show that $\lim_{n\rightarrow \infty}\sum_{k=1}^n\frac{k}{n}a_k=0$. So choose $\varepsilon>0$. Since $\sum_{k=1}^{\infty}a_k\in \mathbb{R}$, there exists an $M$ such that $|a_k|<M$ for all $k$. We can take an $N\in \mathbb{N}$ large enough such that $\sum_{k=1}^{\lfloor \sqrt{N}\rfloor}\frac{\lfloor \sqrt{N}\rfloor}{N}M<\varepsilon$ and $\sum_{k=\lfloor \sqrt{N}\rfloor+1}\frac{k}{N}a_k<\varepsilon$. (Why does such an $N$ exist?)
Now let $n\geq N$, then $$|\sum_{k=1}^n\frac{k}{n}a_k|=|\sum_{k=1}^{\lfloor \sqrt{N}\rfloor}\frac{k}{n}a_k+\sum_{k=\lfloor \sqrt{N}\rfloor+1}^n\frac{k}{n}a_k|\leq |\sum_{k=1}^{\lfloor \sqrt{N}\rfloor}\frac{k}{n}a_k|+|\sum_{k=\lfloor \sqrt{N}\rfloor+1}^n\frac{k}{n}a_k|.$$
Clearly $|\sum_{k=1}^{\lfloor \sqrt{N}\rfloor}\frac{k}{n}a_k|\leq |\sum_{k=1}^{\lfloor \sqrt{N}\rfloor}\frac{\lfloor \sqrt{N}\rfloor}{N}M|$ and $|\sum_{k=\lfloor \sqrt{N}\rfloor+1}^n\frac{k}{n}a_k|<|\sum_{k=\lfloor \sqrt{N}\rfloor+1}^na_k|$.
Putting everything together yields the answer.
