Show $\sum_{n=1}^\infty \dfrac{a_n}{n^s} = l\zeta(s) + o(\zeta(s))$ where $\lim \dfrac{1}{n}\sum_{k=1}^n a_n= l$ Let $(a_n)$ be a real sequence. We suppose thar $\lim \dfrac{1}{n}\sum_{k=1}^n a_n = l \in \Bbb R$.
Show that as $s \rightarrow 1^+$, $$\sum_{n=1}^\infty \dfrac{a_n}{n^s} = l\zeta(s) + o(\zeta(s))$$
Where $\zeta(s) = \sum_{n=1}^\infty \dfrac{1}{n^s}$

What I did : 
I showed that the serie $\sum a_n$ converges using summation by parts.
What I tried :
I don't know how to start

Abel's sum formula is very helpful in matters concerning Dirichlet series. Let
$$A(x) = \sum_{n \leqslant x} a_n,$$
then Abel's sum formula gives
$$\sum_{n \leqslant x} \frac{a_n}{n^s} = \frac{A(x)}{x^{s}} + s\int_1^x \frac{A(t)}{t^{s+1}}\,dt$$
and for $s > 1$ we obtain
$$\sum_{n = 1}^{\infty} \frac{a_n}{n^s} = s\int_1^{\infty} \frac{A(t)}{t^{s+1}}\,dt\,.$$
The assumption on the $a_n$ says that $A(x) = l\cdot x + o(x)$, and therefore
$$\sum_{n = 1}^{\infty} \frac{a_n}{n^s} = s\int_1^{\infty} \frac{l\cdot t + o(t)}{t^{s+1}}\,dt = ls\int_1^{\infty} \frac{dt}{t^s} + s\int_1^{\infty} \frac{o(1)}{t^s}\,dt = \frac{ls}{s-1} + o\bigl(\lvert s-1\rvert^{-1}\bigr)\,.$$
Since $\zeta(s) = \frac{s}{s-1} + O(1)$ as $s \to 1$, the assertion follows.
