# 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 $n > M$:

 $\ds \cmod {1 - \frac z {n^c} }$ $>$ $\ds 1 - \cmod {\frac z {n^c} }$ Triangle Inequality $\ds$ $>$ $\ds 1 - \cmod {\frac R {M^c} }$ $\ds$ $>$ $\ds 1 - \cmod {\frac R {2 R} }$ $\ds$ $=$ $\ds \frac 1 2$ $\ds \frac 1 {\cmod {1 - \frac z {n^c} } }$ $<$ $\ds 2$

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

For all $K > M$:

 $\ds \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c - z} } }$ $=$ $\ds \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} \frac {n^c} {n^c - z} } }$ $\ds$ $=$ $\ds \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} } \cmod {\frac 1 {1 - \frac z {n^c} } } }$ $\ds$ $<$ $\ds \cmod {2 \sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} } }$ $\text {(1)}: \quad$ $\ds$ $<$ $\ds \cmod {2 R \sum_{n \mathop = K}^\infty \cmod {\frac {a_n} {n^c} } }$

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 $K > N$:

 $\ds \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c - z} } }$ $<$ $\ds \cmod {2 R \sum_{n \mathop = K}^\infty \cmod {\frac {a_n} {n^c} } }$ $\ds$ $<$ $\ds \epsilon$

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 $\cmod z < 1$:

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

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 $\cmod z < 1$:

 $\ds \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}$ $=$ $\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} \sum_{k \mathop = 1}^\infty \paren {\frac {\cmod z} {n^c} }^k$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\frac {\cmod z} {n^c} } {1 - \frac {\cmod z} {n^c} }$ Sum of Infinite Geometric Sequence $\text {(2)}: \quad$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {n^c - \cmod z}$

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:

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

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.

$\blacksquare$