Show that $\sum_{n\geq 1} \frac{a_n}{n^z}$ defines a holomorphic function in $\{ \operatorname{Re}z > 1\}$

Question

I am trying to make the following demonstration: let $(a_n)_n\subset \mathbb{C}$ be a sequence fulfilling that for all $\delta>0$, $$\sup_n \dfrac{|a_n|}{n^\delta}<+\infty. $$ Show that the series $\displaystyle \sum_{n\geq 1} \dfrac{a_n}{n^z}$, $\textrm{Re}\, z>1 $, defines a holomorphic function on the half-plane $\textrm{Re}\, z> 1 $.

What I have done is: I first show that $\displaystyle \sum_{n\geq 1} \dfrac{a_n}{n^z}$ converges uniformly in every compact rectangle $R=[a, b] \times [c, d] \subset \mathbb {C} $ with $ a> 1 $. For this I will see that the sequence of partial sums is a uniformly Cauchy sequence. Let $ \beta> \alpha $, and $ z = x + iy \in R $ with $ x> 1 $

\begin{align*} \left| \sum_{n=1}^\beta \dfrac{a_n}{n^z}-\sum_{n=1}^\alpha \dfrac{a_n}{n^z}\right| &=\left| \sum_{n=\alpha +1}^\beta \dfrac{a_n}{n^z}\right|\leq \sum_{n=\alpha+1}^\beta \dfrac{|a_n|}{|n^z|}=\sum_{n=\alpha+1}^\beta \dfrac{|a_n|}{n^x} \\ & \leq \sum_{n=\alpha+1}^\beta \dfrac{|a_n|}{n^a} \leq \sum_{n=\alpha+1}^\beta M=M(\beta -\alpha) \end{align*}

where $M=\displaystyle \sup_{n\in \{ \alpha+1,\ldots, \beta\}} \dfrac{|a_n|}{n^a}$.

If we see that there exists an integer $m$ such that if $m$ exists, then $m\to 0$ then we will have that the succession of partial sums is Cauchy uniform. This is true because as we increase $n$, $n^a$ will increase.

Therefore $\displaystyle \sum_{n \geq 1} \dfrac{a_n}{n^z}$ converges uniformly on every compact subset in the interior of the half-plane of convergence $\textrm{Re} \, z> 1 $ .

Then, $ f (z) = \displaystyle \sum_ {n \geq 1} \dfrac{a_n}{n ^ z} $ is holomorphic in the half-plane $ \textrm {Re} \, z> 1 $ and the sequence of derivatives $ (f'_n) _n $ where $ f_n (z) = \dfrac {a_n} {n ^ z} $ converges uniformly on each compact subset of the half-plane of convergence to the derivative $ f '$ obtained in deriving term by term .

I don't know if it is well resolved. What I don't know how to justify very well is the step that $M$ goes to $0$ for a sufficiently large $N$.

Answer 1

Your estimate $$ \left| \sum_{n=1}^\beta \frac{a_n}{n^z}-\sum_{n=1}^\alpha \frac{a_n}{n^z}\right| \le M(\beta - \alpha) $$ is not good enough:

• $M=\sup_{\alpha+1 \le n \le \beta} \frac{|a_n|}{n^a}$ does not necessarily become small for $\alpha \to \infty$, and
• the factor $\beta - \alpha$ is not bounded independently of $\beta$.

---

One can proceed as follows: For $z = x+iy$ with $x > 1+2 \delta$ and $\delta > 0$ is $$ M = \sup_n \frac{|a_n|}{n^{\delta}} < \infty $$ and therefore $$ \left| \frac{a_n}{n^z}\right| = \frac{|a_n|}{n^x} \le \frac{|a_n|}{n^{1+2\delta}} \le M \frac{1}{n^{1+\delta}} \, . $$ Since $\sum_{n=1}^\infty 1/n^{1+\delta}$ is convergent, it follows with the Weierstrass M-test that the series $$ \sum_{n=1}^\infty \frac{a_n}{n^z} $$ is absolutely and uniformly convergent, and therefore defines a holomorphic function in the halfplane $\{ z \mid \operatorname{Re} z > 1 + 2 \delta \}$.

This holds for all $\delta > 0$, so that the desired conclusion follows.