Analytic Continuation of Generating Function of Dirichlet Series
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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.
$\blacksquare$