Convergence of Dirichlet Series with Bounded Partial Sums

Theorem

Let $\{ a_n \}_{n \in \N}$ be a sequence.

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

$\displaystyle \left| \sum_{k = m}^n a_n \right| \leq B$

Then the Dirichlet series $\displaystyle f(s) = \sum_{n \geq 1} a_n n^{-s}$ converges locally uniformly to an analytic function on $\Re(s) > 0$.

Proof

By Exponential is Entire, the partial sums

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

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

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

Then for any $s \in \C$ such that $\Re(s) > 0$, 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

$\displaystyle f(s) = \sum_{n \geq 1} a_n e^{-\lambda_n s}$

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

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

Let $\epsilon > 0$ be arbitrary, and choose $N_0 \in \N$ such that $\displaystyle \left|e^{-\lambda_n s}\right| < \epsilon$ for all $n \geq N_0$.

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

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert e^{-\alpha s} - e^{-\beta s} \right\vert\)	\(=\)	\(\displaystyle \left\vert s \int_\alpha^\beta e^{-xs}\ dx \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \vert s\vert \int_\alpha^\beta e^{-x \sigma}\ dx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Where $\sigma = \Re(s)$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{\vert s\vert}{\sigma} \left( e^{-\alpha \sigma} - e^{-\beta \sigma} \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Therefore,

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

Finally letting $\Im(s) = t$ we have

$\displaystyle \frac{|s|}{\sigma} = \frac{\sqrt{\sigma^2 + t^2}}{\sigma} = \sqrt{1 + \frac{t^2}{\sigma^2}}$

and

$\displaystyle \frac{\pi}2 > A \geq \arg(s) = \operatorname{atan}\left( \frac t\sigma \right)$

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

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

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

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

$\blacksquare$