# Analytic Continuation of Generating Function of Dirichlet Series

## Theorem

Let $\displaystyle \lambda\left({s}\right) = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series

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

Let $\displaystyle g\left(z\right) = \sum_{k\mathop=1}^\infty \lambda\left(ck\right)z^k$ be the generating function of $\lambda\left(ck\right)$

Then the analytic continuation of $g$ to $\C$ is equal to

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

## Proof

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

For $\left\vert z \right\vert< R$, Pick M such that $M^c > 2R$. This is always possible, as $c > 0$.

Then for $n > M$:

 $\displaystyle \left\vert 1 - \frac {z} {n^c} \right\vert$ $>$ $\displaystyle 1 - \left\vert \frac {z} {n^c} \right\vert$ Triangle Inequality $\displaystyle$ $>$ $\displaystyle 1 - \left\vert \frac {R} {M^c} \right\vert$ $\displaystyle$ $>$ $\displaystyle 1 - \left\vert \frac {R} {2R} \right\vert$ $\displaystyle$ $=$ $\displaystyle \frac {1} {2}$ $\displaystyle \frac 1 {\left\vert 1 - \frac {z} {n^c} \right\vert}$ $<$ $\displaystyle 2$

Therefore, for $\left\vert{z}\right\vert < R$, pick $M$ as described above.

For all $K > M$:

 $\displaystyle \left\vert \sum_{n\mathop=K}^\infty \left\vert a_n \frac {z} {n^c - z } \right\vert \right\vert$ $=$ $\displaystyle \left\vert \sum_{n\mathop=K}^\infty \left\vert a_n \frac {z} {n^c} \frac {n^c} {n^c - z } \right\vert \right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert \sum_{n\mathop=K}^\infty \left\vert a_n \frac {z} {n^c} \right\vert \left\vert \frac {1} {1 - \frac {z} {n^c} } \right\vert \right\vert$ $\displaystyle$ $<$ $\displaystyle \left\vert 2 \sum_{n\mathop=K}^\infty \left\vert a_n \frac {z} {n^c} \right\vert \right\vert$ $(1):\quad$ $\displaystyle$ $<$ $\displaystyle \left\vert 2 R \sum_{n\mathop=K}^\infty \left\vert \frac {a_n} {n^c} \right\vert \right\vert$

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 $K > N$:

 $\displaystyle \left\vert \sum_{n\mathop=K}^\infty \left\vert a_n \frac {z} {n^c - z } \right\vert \right\vert$ $<$ $\displaystyle \left\vert 2 R \sum_{n\mathop=K}^\infty \left\vert \frac {a_n} {n^c} \right\vert \right\vert$ $\displaystyle$ $<$ $\displaystyle \epsilon$

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

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

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 $\left\vert z \right\vert < 1$:

 $\displaystyle \displaystyle g\left(z\right)$ $=$ $\displaystyle \sum_{k\mathop=1}^\infty \lambda\left(ck\right)z^k$ $\displaystyle$ $=$ $\displaystyle \sum_{k\mathop=1}^\infty z^k \sum_{n\mathop=1}^\infty \frac {a_n} {n^{ck} }$ $\displaystyle$ $=$ $\displaystyle \sum_{k\mathop=1}^\infty \sum_{n\mathop=1}^\infty a_n \left(\frac {z} {n^c} \right)^k$

By Fubini-Tonelli Theorem, if:

$\displaystyle \sum_{n\mathop=1}^\infty \sum_{k\mathop=1}^\infty \left\vert a_n \left(\frac {z} {n^c} \right)^k \right\vert$

converges, then:

$\displaystyle \sum_{k\mathop=1}^\infty \sum_{n\mathop=1}^\infty a_n \left(\frac {z} {n^c} \right)^k = \sum_{n=1}^\infty \sum_{k=1}^\infty a_n \left(\frac {z} {n^c} \right)^k$

We see that for $\left\vert z \right\vert<1$:

 $\displaystyle \sum_{n\mathop=1}^\infty \sum_{k\mathop=1}^\infty \left\vert a_n \left(\frac {z} {n^c} \right)^k \right\vert$ $=$ $\displaystyle \sum_{n\mathop=1}^\infty \left\vert a_n \right\vert \sum_{k\mathop=1}^\infty \left(\frac {\left\vert z \right\vert} {n^c} \right)^k$ $\displaystyle$ $=$ $\displaystyle \sum_{n\mathop=1}^\infty \left\vert a_n \right\vert \frac { \frac {\left\vert z \right\vert} {n^c} } {1-\frac {\left\vert z \right\vert} {n^c} }$ Geometric Series $(2):\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{n\mathop=1}^\infty \left\vert a_n \right\vert \frac { \left\vert z \right\vert } {n^c - \left\vert z \right\vert }$

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

$\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{a_n}\right\vert \frac {\left\vert{z}\right\vert } {\left\vert{n^c - z}\right\vert}$

for all but finitely n.

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

Thus we may write:

 $\displaystyle \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \left({\frac z {n^c} }\right)^k$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \left({\frac z {n^c} }\right)^k$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty a_n \frac {\frac z {n^c} } {1 - \frac z {n^c} }$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty a_n \frac z {n^c - z}$

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.

$\blacksquare$