Is $\sum_{j=1}^nja_j=o(n)$ as $n\to\infty$? 
Suppose that $a_j$ are non-negative real numbers such that
  $\sum_{j=1}^\infty a_j<\infty$. Is it true that
  $$
\sum_{j=1}^nja_j=o(n)
$$
  as $n\to\infty$?

I am not sure if it is true or not. It is straightforward to show that $\sum_{j=1}^nja_j=O(n)$ as $n\to\infty$. We have that
$$
\sum_{j=1}^nja_j\le n\sum_{j=1}^na_j\le n\sum_{j=1}^\infty a_j
$$
for each $n\ge 1$. However, the bound is trivial and maybe there is a sharper one.
Any help is much appreciated!
 A: Let $A_k=\sum_{j=1}^{k}a_j$. By summation by parts:
$$ \sum_{j=1}^{n}ja_j = (n+1) A_n - \sum_{j=1}^{n} A_j. \tag{1}$$
For any $\varepsilon>0$, let $n_\varepsilon$ the least natural number such that $\forall n\geq n_\varepsilon$, $\left|A_n-\lim_{n\to +\infty} A_n\right|\leq\varepsilon.$
Let $n=k n_\varepsilon$ and $C=\lim_{n\to +\infty}A_n$: by $(1)$ we have
$$ \sum_{j=1}^{n} ja_j \leq n\varepsilon+ (n_\varepsilon+1) (C+\varepsilon)+\sum_{j=1}^{n_\varepsilon}A_j \tag{2}$$
hence for any $n$ big enough we have $\sum_{j=1}^{n}ja_j \leq 2\varepsilon n$. Since $\varepsilon$ is arbitrary, the claim follows.
A: This is Did's comment posted as an expanded answer (this was Did's suggestion).
Let us denote $A_j=\sum_{k=j}^\infty a_k$ and let us observe that $A_j\to0$ as $j\to\infty$ because $\sum_{j=0}^\infty a_j<\infty$. Since $a_j$'s are non-negative real numbers, we have that
$$
\sum_{j=1}^nja_j\le\sum_{j=1}^nja_j+nA_{n+1}=\sum_{j=1}^nA_j.
$$
For each $\varepsilon>0$, there exists $k(\varepsilon)\ge1$ such that $A_j\le\varepsilon$ for each $j\ge k(\varepsilon)$. For each $n>k(\varepsilon)$, we obtain
$$
\sum_{j=1}^nja_j\le\sum_{j=1}^nA_j=\sum_{j=1}^{k(\varepsilon)}A_j+\sum_{j=k(\varepsilon)+1}^nA_j\le k(\varepsilon)A_1+(n-k(\varepsilon))\varepsilon.
$$
This implies that
$$
\limsup_{n\to\infty}\frac{\sum_{j=1}^nja_j}n\le\varepsilon
$$
for each $\varepsilon>0$. Hence, $\sum_{j=1}^nja_j=o(n)$ as $n\to\infty$.
