how to show that this complex series converge?

Question

If $$\sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}}$$ Converges( s is real) and $\operatorname{Re}(z)>s$. Then $$\sum_{n=1}^{\infty} \frac{a_{n}}{n^{z}}$$ also converges. $a_n$ is complex sequence.

Answer 1

You've got that:

$$\left | \dfrac{a_n}{n^{x+iy}} \right | = \left | \dfrac{a_n}{n^x e^{iy\ln n}} \right | = \dfrac{|a_n|}{n^x | e^{iy \ln(n)}|} = \dfrac{|a_n|}{n^x} \leq \dfrac{|a_n|}{n^s}$$

So the series converges by the comparison test.

Answer 2

You can use ratio test to determine whether the series converges or not. Consider the following series $$ \sum_{n=1}^\infty b_n $$ then $$ \rho=\lim_{n\to\infty}\left|\frac{b_{n+1}}{b_n}\right|. $$ If $\rho<1$, then the series converges. In your case we have $$ \sum_{n=1}^\infty\frac{a_n}{n^s}\;\Rightarrow\;\text{convergent series,} $$ therefore $$ \rho_1=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\left(\frac{n}{n+1}\right)^s\right|<1. $$ Since term $\left(\dfrac{n}{n+1}\right)^s<1$ is always less than $1$ for $s\in\mathbb{Z}_+$, then we can conclude $\left|a_{n+1}\right|<\left|a_{n}\right|$. For the second series we have $$ \rho_2=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\left(\frac{n}{n+1}\right)^z\right|. $$ Since $\left|a_{n+1}\right|<\left|a_{n}\right|$, then term $\left|\dfrac{a_{n+1}}{a_{n}}\right|<1$. Also $|z|>s$, then it is obvious $\left(\dfrac{n}{n+1}\right)^{|z|}<1$. Thus, $\rho_2<1$ and consequently, the second series is also convergent.

$$\\$$

---

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

Answer 3

We can, without loss of generality, assume $s = 0$ (otherwise replace $a_n$ by $b_n = a_n\cdot n^{-s}$).

We prove the convergence of $\sum\limits_{n=1}^\infty \frac{a_n}{n^z}$ for $\operatorname{Re} z > 0$ under the slightly weaker assumption that the sequence

$$P_n = \sum_{k=1}^n a_k$$

of partial sums is bounded.

To prove the convergence of $\sum\limits_{n=1}^\infty \frac{a_n}{n^z}$ for $\operatorname{Re} z > 0$, we check the Cauchy-criterion:

$$\begin{align} \left\lvert \sum_{n=K}^M \frac{a_n}{n^z}\right\rvert &= \left\lvert \sum_{n=K}^M \frac{P_n - P_{n-1}}{n^z}\right\rvert\\ &= \left\lvert \frac{P_M}{M^z} - \frac{P_{K-1}}{K^z} + \sum_{n=K}^{M-1} P_n\left(\frac{1}{n^z}-\frac{1}{(n+1)^z}\right)\right\rvert\\ &\leqslant \frac{\lvert P_M\rvert}{M^x} + \frac{\lvert P_{K-1}\rvert}{K^x} + \sum_{n=K}^{M-1} \lvert P_n\rvert \left\lvert\frac{1}{n^z} - \frac{1}{(n+1)^z} \right\rvert, \end{align}$$

where $x = \operatorname{Re} z$. Now

$$\frac{1}{n^z} - \frac{1}{(n+1)^z} = z\int_{n}^{n+1} \frac{dt}{t^{z+1}}$$

and hence

$$\left\lvert \frac{1}{n^z} - \frac{1}{(n+1)^z}\right\rvert \leqslant \lvert z\rvert \int_n^{n+1} \left\lvert \frac{1}{t^{z+1}}\right\rvert\,dt = \lvert z\rvert \int_{n}^{n+1} \frac{dt}{t^{x+1}}.$$

We have

$$\int_{K}^\infty \frac{dt}{t^{x+1}} = \frac{1}{xK^x},$$

and so finally

$$\left\lvert \sum_{n=K}^M \frac{a_n}{n^z}\right\rvert \leqslant \left(2 + \frac{\lvert z\rvert}{x} \right)\frac{1}{K^x}\cdot \sup_{n} \lvert P_n\rvert.$$

Since $K^{-x}$ converges to $0$ for $K\to \infty$ and $\sup\limits_n \lvert P_n\rvert < \infty$, it follows that

$$s_K = \sum_{n=1}^K \frac{a_n}{n^z}$$

defines a Cauchy sequence, i.e.

$$ \sum_{n=1}^\infty \frac{a_n}{n^z}$$

converges. (The convergence is locally uniform.)

If $\sum a_n$ converges absolutely, the absolute convergence of $\sum \frac{a_n}{n^z}$ for $\operatorname{Re} z \geqslant 0$ follows from the comparison

$$\left\lvert \frac{a_n}{n^z}\right\rvert = \frac{\lvert a_n\rvert}{n^{x}} \leqslant \lvert a_n\rvert,$$

and for bounded $a_n$, the absolute convergence of $\sum \frac{a_n}{n^z}$ for $\operatorname{Re} z > 1$ follows by comparison with $\frac{1}{n^x}$.

However, the example $a_n = (-1)^{n-1}$, where the sequence of partial sums is bounded but not convergent (the Dirichlet $\eta$-function), shows that a Dirichlet series can converge in a strictly larger domain than it converges absolutely. The abscissa of convergence and of absolute convergence can differ by $1$, in contrast to power series, where the convergence is absolute in the open disk of convergence, and conditional convergence can only occur on the boundary of the disk.