Analytic Continuation of Generating Function of Dirichlet Series

Theorem
Let $\ds \map \lambda s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series

Let $c \in \R$ be greater than the abscissa of absolute convergence of $\lambda$ and greater than $0$.

Let $\ds \map g z = \sum_{k \mathop = 1}^\infty \map \lambda {c k} z^k $ be the generating function of $\map \lambda {c k}$

Then the analytic continuation of $g$ to $\C$ is equal to:
 * $\ds \sum_{n \mathop = 1}^\infty a_n \frac z {n^c - z}$

Proof
We will first show that the series is meromorphic functions on $\C$ with simple poles at $n^c$.

For $\cmod z < R$, pick M such that $M^c > 2 R$.

This is always possible, as $c > 0$.

Then for $n > M$:

Therefore, for $\cmod z= < R$, pick $M$ as described above.

For all $K > M$:

Because $c$ is chosen to be greater than the abscissa of absolute convergence of $\lambda$, the above series converges.

Therefore, for all $\epsilon > 0$ we can pick $N$ such that for $K > N$:

Because the summands are meromorphic functions, the series is an absolutely uniformly convergent series of analytic functions on the region $\cmod z < R, z \ne k^c$ for all $k \in \N$.

By Uniform Limit of Analytic Functions is Analytic, the series converges to an analytic function on the domain $z \in \C, z \ne k^c$ for all $k \in \N$.

Also, each $k^c$ is a pole at exactly one summand, where it is a simple pole.

Therefore, the series has simple poles at each $k^c$, meaning that the function is meromorphic on $\C$.

We now show that the Generating Function is equal to the series for $\cmod z < 1$:

By Fubini-Tonelli Theorem, if:


 * $\ds \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}$

converges, then:


 * $\ds \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k = \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k$

We see that for $\cmod z < 1$:

The summands of $(2)$ are equal to the summands of:


 * $\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {\cmod {n^c - z} }$

for all but finitely n.

By $(1)$, the above series converges, and thus $(2)$ as well.

Thus we may write:

Thus the Generating function is equal to the series on the unit circle.

Thus by Definition:Analytic Continuation, the series is the unique analytic continuation of the Generating Function.