# 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.

• Are the $a_n$ arbitrary complex numbers, or have we some helpful constraints like $a_n \geqslant 0$? – Daniel Fischer May 13 '14 at 19:52
• No. $a_n$ arbitrary complex numbers – ZFR May 13 '14 at 20:00
• Can anyone hlp with this problem? – ZFR May 14 '14 at 12:21

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 $n \geqslant n_\varepsilon$.