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
 A: 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.
A: It suffices to prove that $$\forall \epsilon, \exists \delta, s\in (1,1+\delta)\implies \left| \sum_{n=1}^\infty \frac{a_n}{n^s} -l \sum_{n=1}^\infty \frac{1}{n^s}\right| \leq \epsilon \sum_{n=1}^\infty \frac{1}{n^s}$$
Letting $b_n = \frac{1}{n}\sum_{k=1}^n a_n$, 
$$\sum_{n=1}^\infty \frac{a_n}{n^s} -l \sum_{n=1}^\infty \frac{1}{n^s}  = \sum_{n=1}^\infty\left( \frac{b_n-b_{n-1}}{n^{s-1}} + \frac{b_{n-1}-l}{n^s}\right)$$
Summation by part yields $\displaystyle \sum_{n=1}^\infty\frac{b_n-b_{n-1}}{n^{s-1}} = \sum_{n=1}^\infty b_n\left(\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}} \right)$.
By the mean value theorem, for each $n$, there is some $x_n\in (n,n+1)$ such that $$\frac{1}{n^{s-1}}-\frac{1}{(n+1)^{s-1}} = \frac{s-1}{x_n^{s}}$$
Thus $$\left| \sum_{n=1}^\infty \frac{a_n}{n^s} -l \sum_{n=1}^\infty \frac{1}{n^s}\right| \leq \sum_{n=1}^\infty\left| \frac{b_n(s-1)}{x_n^{s}}\right| + \sum_{n=1}^\infty \left|\frac{b_{n-1}-l}{n^s}\right|$$

Since $b_n$ converges, it is bounded by some $M$, and $x_n> n$, thus $$\sum_{n=1}^\infty\left| \frac{b_n(s-1)}{x_n^{s}}\right|\leq M(s-1)\sum_{n=1}^\infty \frac 1{n^s} $$
Using integrals, one easily shows that $\zeta(s)\leq \frac{1}{s-1} +1$, hence $$\sum_{n=1}^\infty\left| \frac{b_n(s-1)}{x_n^{s}}\right|\leq Ms$$

Since $b_n$ converges to $l$, there is some $N$ such that $n\geq N\implies |b_n-l|\leq \epsilon$.
Then $$\sum_{n=1}^\infty \left|\frac{b_{n-1}-l}{n^s}\right|\leq \sum_{n=1}^{N}\left|\frac{b_{n-1}-l}{n^s}\right| +\epsilon\sum_{n=N+1}^{\infty}\frac{1}{n^s}= \sum_{n=1}^{N}\frac{\left|b_{n-1}-l\right|-1}{n^s} +\epsilon\sum_{n=1}^{\infty}\frac{1}{n^s} $$

To finish the proof, note that $\lim_{s\to 1^+} \zeta(s) = \infty$, so the bound $$\left| \sum_{n=1}^\infty \frac{a_n}{n^s} -l \sum_{n=1}^\infty \frac{1}{n^s}\right|\leq \underbrace{Ms + \sum_{n=1}^{N}\frac{\left|b_{n-1}-l\right|-1}{n^s}}_{\text{has a finite limit}} +\epsilon\sum_{n=1}^{\infty}\frac{1}{n^s}$$
does the job.
