Series divergence 
Prove that if $\sum a_n$ is a convergent series of positive terms  then the series $\sum b_n$ is always divergent where $b_n = \frac{a_1+ a_2+\dots+a_n}{n}$ for every integer $n$.

I tried to solve it by looking at the partial sums of $b_n$, comparison test, ratio test, but nothing seemed to work out.
Any help is appreciated!
 A: A famous result is that if $x_n\to \ell$, then $$\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n x_n=\ell.$$
Therefore, if $$\lim_{n\to \infty }\sum_{k=1}^n a_n=\ell>0,$$
then obviously $$\sum_{k=1}^\infty b_n=+\infty .$$
And if $\sum_{k=1}^\infty a_n=0$, since $a_n\geq 0$, you must have $a_n=0$ for all $n$, and then $\sum_{n}b_n=0$, and thus it converge.
A: It suffices to know that the sequence $\{a_n\}_{n\geq 1}$ is non-negative with at least one positive term $a_N$ (we need not the convergence of $\sum_n a_n$). Then $b_n$ is non-negative too and for $n\geq N$ 
$$b_n= \frac{a_1+ a_2+\dots +a_n}{n}\geq \frac{a_N}{n}.$$
It follows that
$$\sum_{n=1}^{\infty} b_n\geq \sum_{n=N}^{\infty} b_n\geq a_N\sum_{n=N}^{\infty} \frac{1}{n}=+\infty.$$
A: Note that
$$b_n=\frac{\sum_{k=1}^n a_k}{n}$$
which diverges by limit comparison test with $c_n=\frac1n$ indeed 
$$\frac{b_n}{c_n}=\sum_{k=1}^n a_k=L\in \mathbb{R}$$
A: Let $\sum_{n=1}^{\infty} a_n=S$. Then:
$$\sum_{n=1}^{\infty} b_n\ge \begin{cases} \sum_{n=1}^{\infty} \frac{1}{Sn}=+\infty, \ S\ge 1 \\ \sum_{n=1}^{\infty} \frac{S}{n}=+\infty, \ 0<S<1\end{cases}$$
