<s>[INST] Dirichlet series Suppose that the series $\sum \limits_{n=1}^{\infty}\dfrac{a_n}{n^{\sigma}} \quad(\sigma>0)$ converges. Prove that $\lim \limits_{n\to \infty}\dfrac{a_1+a_2+\dots+a_n}{n^{\sigma}}=0$.
I applied the Abel transformation, but unsuccessfully.
 [/INST] For each $n$, let $R_n = \sum\limits_{k=n}^\infty \frac{a_k}{k^\sigma}$. Then $a_k = k^\sigma(R_k-R_{k+1})$. Since by assumption the series converges, we have $R_n \to 0$, thus $S_n = \sup\limits_{k\geqslant n} \lvert R_k\rvert$ is finite for all $n$ and $S_n \to 0$. Then we have
$$\begin{align}
\left\lvert \frac{a_1 + \dotsc + a_n}{n^\sigma}\right\rvert
&= n^{-\sigma}\left\lvert \sum_{k=1}^n k^\sigma(R_k - R_{k+1})\right\rvert\\
&= n^{-\sigma}\left\lvert \sum_{k=1}^n k^\sigma R_k - \sum_{k=2}^{n+1} (k-1)^\sigma R_k\right\rvert\\
&= n^{-\sigma}\left\lvert R_1 - n^\sigma R_{n+1} + \sum_{k=2}^n \left(k^\sigma - (k-1)^\sigma\right)R_k\right\rvert\\
&\leqslant \frac{S_1}{n^\sigma} + S_{n+1} + \sum_{k=2}^n \frac{k^\sigma - (k-1)^\sigma}{n^\sigma}S_k\\
&\leqslant \frac{S_1}{n^\sigma} + S_{n+1} + S_2\sum_{k=2}^{\lfloor\sqrt{n}\rfloor} \frac{k^\sigma - (k-1)^\sigma}{n^\sigma} + S_{\lfloor\sqrt{n}\rfloor+1}\sum_{k=\lfloor\sqrt{n}\rfloor+1}^n\frac{k^\sigma-(k-1)^\sigma}{n^\sigma}\\
&= \frac{S_1}{n^\sigma} + S_{n+1} + S_2\frac{\lfloor\sqrt{n}\rfloor^\sigma-1}{n^\sigma} + S_{\lfloor\sqrt{n}\rfloor+1} \frac{n^\sigma - \lfloor\sqrt{n}\rfloor^\sigma}{n^\sigma}\\
&\leqslant \frac{S_1}{n^\sigma} + S_{n+1} + \frac{S_2}{n^{\sigma/2}} + S_{\lfloor\sqrt{n}\rfloor+1}.
\end{align}$$
Now, given $\varepsilon > 0$, one can choose $n_\varepsilon$ so large that $S_{\lfloor\sqrt{n_\varepsilon}\rfloor+1} < \frac{\varepsilon}{4}$ and $\frac{S_1}{n_\varepsilon^{\sigma/2}} < \frac{\varepsilon}{4}$. Then, one has
$$\left\lvert\frac{a_1+\dotsc + a_n}{n^\sigma}\right\rvert < \varepsilon$$
for all $n \geqslant n_\varepsilon$.
</s>