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 $\cmod z= < R$, 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 $\cmod z < R, z \ne k^c \text{ for all } k \in \N$. By Uniform Limit of Analytic Functions is Analytic, the series converges to an analytic function on the domain $z \in \C, z \ne k^c \text{ for all } k \in \N$.  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 $\cmod z < 1$: {{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: :$\displaystyle \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \cmod {a_n \paren {\frac z {n^c} }^k}$ converges, then: :$\displaystyle \sum_{k \mathop = 1}^\infty \sum_{n \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k = \sum_{n \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty a_n \paren {\frac z {n^c} }^k$ 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: :$\displaystyle \sum_{n \mathop = 1}^\infty \cmod {a_n} \frac {\cmod z} {\cmod {n^c -  z} }$ 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 55n7uoh84aokhcwg5f0uin4d4rvcscy