Dirichlet Series Convergence Lemma/General

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Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \left({s}\right)}$ be a general Dirichlet series.

Let $f \left({s}\right)$ converge at $s_0 = \sigma_0 + i t_0$.

Then $f \left({s}\right)$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.


Let $s=\sigma+it$

Let $s_0$ be such that $f \left({s_0}\right)$ converges.

Let $S\left(m,n\right) = \displaystyle \sum_{k \mathop = n}^m a_k e^{-\lambda_ks_0}$

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

\(\displaystyle g \left({s}\right)\) \(=\) \(\displaystyle f \left({s + s_0}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \left({s + s_0}\right)}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s_0} e^{-\lambda_n s}\) $\quad$ $\quad$

Thus it suffices to show $g \left({s}\right)$ 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 $m,n>N$:

$ \left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert < \epsilon$

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

\(\displaystyle \left\vert \sum_{k \mathop = n}^m a_k e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert\) \(=\) \(\displaystyle \left\vert \sum_{k \mathop = n}^m \left(S\left(k,n\right) - S\left(k-1,n\right)\right) e^{-\lambda_ks} \right\vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\vert S\left(m,n\right)e^{-\lambda_ms} + \sum_{k \mathop = n}^{m-1} S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert\) $\quad$ $\quad$

Because $S\left(m,n\right)$ is the difference of partial sums of a convergent, and thus cauchy, sequence, its modulus, $\left\vert S\left(m,n\right) \right\vert$, is bounded, say by $Q$.

Thus we have:

\(\displaystyle \left\vert S\left(m,n\right)^{-\lambda_ms} + \sum_{k \mathop = n}^{m-1} S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert\) \(\leq\) \(\displaystyle \left\vert S\left(m,n\right) e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert\) $\quad$ $\quad$
\(\displaystyle \) \(\leq\) \(\displaystyle Qe^{-\lambda_m\sigma}+ Q \sum_{k \mathop = n}^{m-1} \left\vert e^{-\lambda_ks} - e^{-\lambda_{k+1}s} \right\vert\) $\quad$ $\quad$

We see that:

\(\displaystyle \left\vert e^{-\lambda_ks} - e^{-\lambda_{k+1}s} \right\vert\) \(=\) \(\displaystyle \left\vert \int_{\lambda_k}^{\lambda_{k+1} } -se^{-xs} \rd x \right\vert\) $\quad$ $\quad$
\(\displaystyle \) \(\leq\) \(\displaystyle \int_{\lambda_k}^{\lambda_{k+1} } \left\vert-se^{-xs}\right\vert \rd x\) $\quad$ Modulus of Complex Integral $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_{\lambda_k}^{\lambda_{k+1} } \left\vert s \right\vert e^{-x\sigma} \rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\vert s \right\vert \int_{\lambda_k}^{\lambda_{k+1} } e^{-x\sigma} \rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left\vert s \right\vert} {\sigma} \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)\) $\quad$ $\quad$

Thus we have:

\(\displaystyle Qe^{-\lambda_m\sigma} + Q \sum_{k \mathop = n}^{m-1} \left\vert e^{-\lambda_ks} - e^{-\lambda_{k+1}s} \right\vert\) \(\leq\) \(\displaystyle Qe^{-\lambda_m\sigma} + Q \sum_{k \mathop = n}^{m-1} \frac {\left\vert s \right\vert} {\sigma} \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle Q e^{-\lambda_m\sigma} + Q \frac {\left\vert s \right\vert} {\sigma} \left( e^{-\lambda_n\sigma} - e^{-\lambda_{m}\sigma} \right)\) $\quad$ Telescoping Series $\quad$

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