Dirichlet Series Convergence Lemma

Theorem

Let $\displaystyle f(s) = \sum_{n = 1}^\infty \frac{a_n}{n^s}$ be a Dirichlet series.

If $f(s)$ converges at $s_0 = \sigma_0 + it_0$, then it converges for all $s=\sigma + it$ with $\sigma > \sigma_0$.

Proof

We begin with a lemma.

Lemma

Let $\displaystyle f(s) = \sum_{n = 1}^\infty \frac{a_n}{n^s}$ be a Dirichlet series.

Suppose that For some $s_0 = \sigma_0 + it_0 \in \C$, $f(s_0)$ has bounded partial sums:

$\displaystyle (1) \qquad \left|{ \sum_{n=1}^N a_n n^{-s_0} }\right| \leq M$

for some $M \in \R$ and all $N \geq 1$.

Then for every $s = \sigma + it \in \C$ with $\sigma > \sigma_0$,

$\displaystyle \left|{ \sum_{n = m}^N a_n n^{-s} }\right| \leq 4 M m^{\sigma_0-\sigma}$

Proof of Lemma

We have the Summation by Parts formula:

$\displaystyle \sum_{n=m}^N f_n g_n = f_N G_N - f_n G_{n-1} - \sum_{n = m}^{N - 1} G_n\left( f_{n+1} - f_n \right)$

We let $g_n = a_n n^{-s_0}$ and $f_n = n^{s_0 - s}$.

For $N \geq 1$, the quantities $G_N$ are the partial sums $(1)$, so $G_N \leq M$ for all $N \geq 1$. We have

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert \sum_{n = m}^N \frac{a_n}{n^s} \right\vert\)	\(=\)	\(\displaystyle \left\vert \sum_{n=m}^N f_n g_n \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \left\vert f_N G_N \right\vert + \left\vert f_n G_{n-1} \right\vert + \sum_{n = m}^{N - 1} \left\vert G_n\left( f_{n+1} - f_n \right) \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	using partial summation and the Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle M \left\vert N^{s_0 - s} \right\vert + M \left\vert m^{s_0 - s} \right\vert + M \sum_{n = m}^{N - 1} \left\vert \left( (n + 1)^{s_0 - s} - n^{s_0 - s} \right) \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	using the given bound on the partial sums
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle M N^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} + M \left\vert N^{s_0 - s} - m^{s_0 - s} \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	telescoping the series on the right
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle 2M m^{\sigma_0 - \sigma} + M m^{\sigma_0 - \sigma} \left\vert \left( \frac{N}{m} \right)^{s_0 - s} - 1 \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	because $\sigma > \sigma_0$ and $N \geq m$

Finally, because $ \displaystyle \frac{N}{m} > 1$ and $\sigma_0-\sigma < 0$, we can estimate

$\displaystyle \left\vert \left( \frac{N}{m} \right)^{s_0 - s} - 1 \right\vert \leq \left( \frac{N}{m} \right)^{\sigma_0-\sigma} + 1 \leq 2$

Therefore, we have

$\displaystyle \left\vert \sum_{n = m}^N \frac{a_n}{n^s} \right\vert\leq 4M m^{\sigma_0 - \sigma}$

The desired result.

$\Box$

Proof of Theorem

Suppose that $f(s)$ converges at $s_0 = \sigma_0 + it_0$, and choose any $s=\sigma + it$ with $\sigma > \sigma_0$.

The lemma shows that for a constant $C$ independent of $m$,

$\displaystyle \left|{ \sum_{n = m}^N a_n n^{-s} }\right| \leq C m^{\sigma_0-\sigma}$

Since $\sigma_0 - \sigma <0$, the left hand side tends to zero as $m \to \infty$.

Therefore, it follows from Cauchy's Convergence Criterion that the sum converges.

Sources

• Tom M. Apostol: Introduction to Analytic Number Theory (1976): $\S 11.6$: Lemma $2$, Theorem $11.8$