# Dirichlet Series Convergence Lemma/General

## Theorem

Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.

Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.

Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.

## Proof

Let $s = \sigma + i t$

Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.

Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$

We may create a new Dirichlet series that converges at 0 by writing:

 $\ds \map g s$ $=$ $\ds \map f {s + s_0}$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \paren {s + s_0} }$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s_0} e^{-\lambda_n s}$

Thus it suffices to show $\map g s$ converges for $\sigma > 0$.

By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$ there exists an $N$ such that for all $m, n > N$:

$\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} } < \epsilon$

By Abel's Lemma: Formulation 2 we may write:

 $\ds \cmod {\sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0} e^{-\lambda_k s} }$ $=$ $\ds \cmod {\sum_{k \mathop = n}^m \paren {\map S {k, n} - \map S {k - 1, n} } e^{-\lambda_k s} }$ $\ds$ $=$ $\ds \cmod {\map S {m, n} e^{-\lambda_m s} + \sum_{k \mathop = n}^{m - 1} \map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }$

Because $\map S {m, n}$ is the difference of partial sums of a convergent, and thus cauchy, sequence, its modulus, $\cmod {\map S {m, n} }$, is bounded, say by $Q$.

Thus we have:

 $\ds \cmod {\map S {m, n}^{-\lambda_m s} + \sum_{k \mathop = n}^{m - 1} \map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }$ $\le$ $\ds \cmod {\map S {m, n} e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m-1} \cmod {\map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }$ $\ds$ $\le$ $\ds Q e^{-\lambda_m \sigma} + Q \sum_{k \mathop = n}^{m - 1} \cmod {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} }$

We see that:

 $\ds \cmod {e^{-\lambda_k s} - e^{-\lambda_{k+1}s} }$ $=$ $\ds \cmod {\int_{\lambda_k}^{\lambda_{k + 1} } -s e^{-x s} \rd x}$ $\ds$ $\le$ $\ds \int_{\lambda_k}^{\lambda_{k + 1} } \cmod {-s e^{-x s} } \rd x$ Modulus of Complex Integral $\ds$ $=$ $\ds \int_{\lambda_k}^{\lambda_{k + 1} } \cmod s e^{-x \sigma} \rd x$ $\ds$ $=$ $\ds \cmod s \int_{\lambda_k}^{\lambda_{k + 1} } e^{-x \sigma} \rd x$ $\ds$ $=$ $\ds \frac {\cmod s} \sigma \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }$

Thus we have:

 $\ds Q e^{-\lambda_m \sigma} + Q \sum_{k \mathop = n}^{m - 1} \cmod {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} }$ $\le$ $\ds Q e^{-\lambda_m \sigma} + Q \sum_{k \mathop = n}^{m - 1} \frac {\cmod s} {\sigma} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }$ $\ds$ $=$ $\ds Q e^{-\lambda_m \sigma} + Q \frac {\cmod s} \sigma \paren {e^{-\lambda_n \sigma} - e^{-\lambda_m \sigma} }$ Telescoping Series

Because $\lambda_n$ tends to infinity, both summands tend to $0$ as $n$ goes to $\infty$ if $\sigma > 0$.

Thus we can pick $N$ large enough such that both terms are less than $\dfrac \epsilon 2$ for $n, m > N$, giving us the desired result.

$\blacksquare$