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 $ka_k$?
 A: 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,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} ka_k=\lim_{n\to\infty}A_n
-\lim_{n\to\infty}\frac{\sum_{k=1}^{n}A_{k-1}}{n}=
A-\lim_{n\to\infty}\frac{A_n}{(n+1)-n}=A-A=0.$$
A: 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.
A: Is is just an application of Kronercker's Lemma 
 https://en.wikipedia.org/wiki/Kronecker%27s_lemma

By simple choosing $b_n=n$, you can get what you want. 
