Convergence of Dirichlet Series with Bounded Partial Sums

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Theorem

Let $\left\langle{a_n}\right\rangle_{n \mathop \in \N}$ be a sequence in $\C$.

Suppose that there exists $B > 0$ such that for all $n, m \in \N$:

$\displaystyle \left\vert{\sum_{k \mathop = m}^n a_n}\right\vert \le B$

Then the Dirichlet series:

$\displaystyle f \left({s}\right) = \sum_{n \mathop \ge 1} a_n n^{-s}$

converges locally uniformly to an analytic function on $\Re \left({s}\right) > 0$.


Proof

By Exponential is Entire, the partial sums:

$\displaystyle f_N \left({s}\right) = \sum_{n \mathop = 1}^N a_n n^{-s}$

are analytic.

So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence.


For $0 < A < \pi / 2$, $\delta > 0$ we let:

$D_{A, \delta} = \left\{{s \in \C : -A < \arg \left({s}\right) < A,\ \Re \left({s}\right) > \delta}\right\}$

Then for any $s \in \C$ such that $\Re \left({s}\right) > 0$, we can choose $A$, $\delta$ such that $s \in D_{A,\delta}$.

So it is sufficient to prove locally uniform convergence in this region.


Also note that if $\lambda_n = \log n$, then:

$\displaystyle f \left({s}\right) = \sum_{n \mathop \ge 1} a_n e^{-\lambda_n s}$


By Abel's Lemma: Formulation 2, for $N, M \in \N$ we have:

$\displaystyle \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} = \sum_{n \mathop = M}^{N - 1} \left({\sum_{k \mathop = M}^n a_n}\right) \left[{e^{-\lambda_n s} - e^{-\lambda_{n + 1} s}}\right] + e^{-\lambda_N s} \left({\sum_{n \mathop = M}^N a_n}\right)$


Let $\epsilon > 0$ be arbitrary, and choose $N_0 \in \N$ such that:

$\displaystyle \forall n \ge N_0: \left|{e^{-\lambda_n s}}\right| < \epsilon$

Now for $N, M \ge N_0$, by the above we obtain:

$(1): \quad \displaystyle \left\vert \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} \right\vert \le B \sum_{n \mathop = M}^{N - 1} \left\vert e^{-\lambda_n s} - e^{-\lambda_{n+1} s} \right\vert + B \epsilon$


For any $\alpha, \beta \in \R$ we have:

\(\ds \left\vert{e^{-\alpha s} - e^{-\beta s} }\right\vert\) \(=\) \(\ds \left\vert{s \int_\alpha^\beta e^{-x s} \ \mathrm d x}\right\vert\)
\(\ds \) \(\le\) \(\ds \left\vert{s}\right\vert \int_\alpha^\beta e^{-x \sigma} \ \mathrm d x\) where $\sigma = \Re \left({s}\right)$
\(\ds \) \(=\) \(\ds \frac{\left\vert{s}\right\vert} \sigma \left({e^{-\alpha \sigma} - e^{-\beta \sigma} }\right)\)

Therefore:

\(\ds \left\vert{\sum_{n \mathop = M}^N a_n e^{-\lambda_n s} }\right\vert\) \(\le\) \(\ds \frac {B \left\vert{s}\right\vert} \sigma \sum_{n \mathop = M}^{N - 1} \left({e^{-\lambda_n \sigma} - e^{-\lambda_{n + 1} \sigma} }\right) + B \epsilon\) by $(1)$
\(\ds \) \(=\) \(\ds \frac {B \left\vert{s}\right\vert} \sigma \left({e^{-\lambda_N \sigma} - e^{-\lambda_M \sigma} }\right) + B \epsilon\)
\(\ds \) \(\le\) \(\ds \frac {B \epsilon \left\vert{s}\right\vert} \sigma + B \epsilon\)

Finally letting $\Im \left({s}\right) = t$:

$\dfrac{\left\vert{s}\right\vert} \sigma = \dfrac{\sqrt{\sigma^2 + t^2} } \sigma = \sqrt{1 + \dfrac {t^2} {\sigma^2} }$

and:

$\dfrac {\pi} 2 > A \ge \arg \left({s}\right) = \arctan \left({\dfrac t \sigma}\right)$

so $\dfrac t \sigma < \tan A$ is bounded uniformly in $D_{A, \delta}$.

Thus letting $N \to \infty$, we have shown that:

$\displaystyle \left\vert{\sum_{n \mathop = M}^\infty a_n e^{-\lambda_n s} }\right\vert \to 0$

as $M \to \infty$ uniformly in $D_{A, \delta}$.

$\blacksquare$