Whether a series is convergent or divergent Is it true that if $\sum u_n$ is convergent, where $u_n$'s are positive real numbers then $\sum \dfrac{u_1+u_2+...+u_n}{n}$ is divergent?
I know that if   $\lim_{n\to\infty}u_n =0$ then $\lim_{n\to\infty}\dfrac{u_1+u_2+...+u_n}{n}=0$ and it is the necessary condition for a series to be convergent.
Someone help please.
 A: Hint. If $u_i\geq 0$ and $u_1>0$ then
$$\sum_{n=1}^N \dfrac{u_1+u_2+...+u_n}{n}\geq \sum_{n=1}^N \dfrac{u_1}{n}.$$
A: If $\sum_{n\geq 1}u_n$ is convergent to some $\ell\neq 0$, for any $\varepsilon>0$ there is some $N_\varepsilon$ ensuring $\left|-\ell+\sum_{n=1}^{N}u_n\right|\leq \varepsilon$ for any $N\geq N_\varepsilon$. In particular, from some point on the terms of the series $\sum_{n\geq 1}\frac{u_1+\ldots+u_n}{n}$ have the same sign and their absolute values are greater than $\frac{\left|\ell\right|}{2n}$. Since $\sum_{n\geq N_\varepsilon}\frac{C}{n}$ is divergent for any $N_\varepsilon\geq 1$ and any $C\neq 0$, so it is the series $\sum_{n\geq 1}\frac{u_1+\ldots+u_n}{n}$ .
A: The post has been edited to say the $u_i$'s are positive.  Under that assumption, $\sum \dfrac{u_1+u_2+...+u_n}{n}$ diverges because $\sum_{n=1}^m \dfrac{u_1+u_2+...+u_n}{n}>\sum_{n=1}^m \dfrac{u_1}{n}$ and the latter goes to infinity as $m\to\infty$.
A: Answering the more general question before the edit, where the $u_i$'s are not necessarily positive.
This will be true under an additional assumption, namely that the series $\sum_n u_n$ converges to a non-zero number. 
Indeed, suppose $\sum_{n=0}^\infty u_n = S\neq 0$ (without loss of generality, assume $S>0$), then we have that (by definition) there exists $N\geq 0$ such that $$\frac{S}{2} < S_n \stackrel{\rm def}{=} \sum_{k=0}^k u_k \leq 2S$$
for all $n\geq N$.
But then,
$$\frac{S_n}{n} \geq \frac{S}{2}\cdot \frac{1}{n}$$
for all $n\geq N$, so the series $\sum_{n}\frac{S_n}{n}$ diverges by comparison with the Harmonic series.
