Theorem: Analytic Continuation of Generating Function of Dirichlet Series

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 $n > M$: 
{{begin-eqn}}
{{eqn | l = \cmod {1 - \frac z {n^c} }
      | o = >
      | r = 1 - \cmod {\frac z {n^c} }
      | c = Triangle Inequality 
}}
{{eqn | o = >
      | r = 1 - \cmod {\frac R {M^c} }
}}
{{eqn | o = >
      | r = 1 - \cmod {\frac R {2 R} }
}}
{{eqn | r = \frac 1 2
}}
{{eqn | l = \frac 1 {\cmod {1 - \frac z {n^c} } }
      | o = <
      | r = 2
}}
{{end-eqn}}
Therefore, for $\cmod z \le R$, pick $M$ as described above.
For all $K > M$:
{{begin-eqn}}
{{eqn | l = \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c - z} } }
      | r = \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} \frac {n^c} {n^c - z} } }
}}
{{eqn | r = \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} } \cmod {\frac 1 {1 - \frac z {n^c} } } }
}}
{{eqn | o = < 
      | r = \cmod {2 \sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c} } }
}}
{{eqn | n = 1
      | o = < 
      | r = \cmod {2 R \sum_{n \mathop = K}^\infty \cmod {\frac {a_n} {n^c} } }
}}
{{end-eqn}}
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 $K > N$:
{{begin-eqn}}
{{eqn | l = \cmod {\sum_{n \mathop = K}^\infty \cmod {a_n \frac z {n^c - z} } }
      | o = < 
      | r = \cmod {2 R \sum_{n \mathop = K}^\infty \cmod {\frac {a_n} {n^c} } }
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
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 $\cmod z < 1$:
{{begin-eqn}}
{{eqn | l = \map g z
      | r = \sum_{k \mathop = 1}^\infty \map \lambda {c k} z^k
}}
{{eqn | r = \sum_{k \mathop = 1}^\infty z^k \sum_{n \mathop = 1}^\infty \frac {a_n} {n^{c k} }
}}
{{eqn | r = \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k
}}
{{end-eqn}}
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 $\cmod z < 1$: 
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}
      | r = \sum_{n \mathop = 1}^\infty \cmod {a_n} \sum_{k \mathop = 1}^\infty \paren {\frac {\cmod z} {n^c} }^k 
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\frac {\cmod z} {n^c} } {1 - \frac {\cmod z} {n^c} }
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | n = 2
      | r = \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {n^c - \cmod z}
}}
{{end-eqn}}
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: 
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k
      | r = \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty a_n \frac {\frac z {n^c} } {1 - \frac z {n^c} }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty a_n \frac z {n^c - z}
}}
{{end-eqn}}
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. 
{{qed}}
Category:Dirichlet Series
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