Convergence Of Dirichlet Series

The Riemann Zeta series is an example of a Dirichlet series.

$$\zeta (s)=\sum \frac{1}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\dots$$

Dirichlet series have the general form $\sum a_n/n^s$, in contrast to the more familiar power series $\sum a_n{z}^n$.

In this blog post we'll explore when these series converge, first looking at absolute convergence, and then more general convergence.

The video for this post is at [youtube], and the slides are here [pdf].

Abscissa Of Absolute Convergence

A series converges absolutely even when all its terms are replaced by their magnitudes, sometimes called absolute values. This is quite a strong condition, and not all series that converge do so absolutely.

Let's assume a Dirichlet series converges absolutely at $s_1={\sigma }_1+i{t}_1$, and consider another point $s_2={\sigma }_2+i{t}_2$ where ${\sigma }_2\ge {\sigma }_1$. On the complex plane, $s_2$ is to the right of $s_1$.

Now let's compare the magnitudes of the terms in this series at $s_1$ and $s_2$. Remember $n^{\sigma +it}=n^{\sigma }{e}^{it\ln n}$, and because the magnitude of any $e^{i\theta }$ is 1, we can simplify $\left|n^{\sigma +it}\right|=n^{\sigma }$.

$$\sum \left|\frac{a_n}{n^{s_1}}\right|=\sum \frac{\left|a_n\right|}{n^{{\sigma }_1}}\ge \sum \frac{\left|a_n\right|}{n^{{\sigma }_2}}=\sum \left|\frac{a_n}{n^{s_2}}\right|$$

This is simply telling us that the magnitude of each term in the series at $s_2$ is less than or equal to the magnitude of the same term at $s_1$. So if the series converges at $s_1$, it must also converge at $s_2$. More generally, the series converges at any $s=\sigma +it$ where $\sigma \ge {\sigma }_1$. 

If our series doesn't converge everywhere, the $s$ for which it diverges must therefore have $\sigma <{\sigma }_1$. We can see there must be a minimum ${\sigma }_a$, called the abscissa of absolute convergence, such that the series converges for all $\sigma >{\sigma }_a$. 

Notice how absolute convergence depends only on the real part of $s$. Working out the domain of convergence along the real line automatically gives us the domain of convergence in the complex plane.

For example, in we previously showed the series $\sum 1/n^{\sigma }$ converges for real $\sigma >1$. We also know the series diverges at $\sigma =1$. These two facts allow us to say ${\sigma }_a=1$, and so the series converges for all complex $s=\sigma +it$ where $\sigma >1$.

It's interesting that the region of convergence for a Dirichlet series is a half-plane, whereas the region for the more familiar power series $\sum a_n{z}^n$ is a circle.

Abscissa Of Convergence

Absolute convergence is easier to explore as we don't need to consider the effect of complex terms which contribute a negative amount to the overall magnitude of the series. For example, a term $e^{i\pi }=-1$ can partially cancel the effect of a term $2e^{i2\pi }=+2$. This cancelling effect can mean some series do converge, even if not absolutely.

Our strategy, inspired by Apostol, will be to show that if a Dirichlet series is bounded at $s_0={\sigma }_0+i{t}_0$ then it is also bounded at $s=\sigma +it$, where $\sigma >{\sigma }_0$, and then push a little further to show it actually converges at that $s$. 

Let's start with a Dirichlet series $\sum a_n/n^s$ that we know has bounded partial sums at a point $s_0={\sigma }_0+i{t}_0$ for all $x\ge 1$. 

$$\left|\sum _{n\le x}\frac{a_n}{n^{s_0}}\right|\le M$$

Being bounded is not as strong a requirement as convergence, the partial sums could oscillate for example.

We'll use Abel's partial summation formula, explained in a later blog, which relates a discrete sum to a continuous integral.

$$\sum _{x_1<n\le x_2}{b}_{n}f(n)=B(x_2)f(x_2)-B(x_1)f(x_1)-{\int }_{x_1}^{x_2}B(t)f^{\prime }(t)dt$$

Because we're comparing to $s_0$, we'll define $f(x)=x^{s_0-s}$ and $b_n=a_n/n^{s_0}$. Here $B(x)$ is defined as $\sum _{n\le x}{b}_n$, and so $|B(x)|\le M$.

$$\begin{array}{rl}\sum _{x_1<n\le x_2}\frac{a_n}{n^s}& =\sum _{x_1<n\le x_2}{b}_{n}f(n)\\ \\ & =\frac{B(x_2)}{x_2^{s-s_0}}-\frac{B(x_1)}{x_1^{s-s_0}}+(s-s_0){\int }_{x_1}^{x_2}\frac{B(t)}{t^{s-s_0+1}}dt\end{array}$$

We now consider the magnitude of the series, which is never more than the sum of the magnitudes of its parts, and make use of $|B(x)|\le M$. 

$$\begin{array}{rl}\left|\sum _{x_1<n\le x_2}\frac{a_n}{n^s}\right|& \le \left|\frac{B(x_2)}{x_2^{s-s_0}}\right|+\left|\frac{B(x_1)}{x_1^{s-s_0}}\right|+|(s-s_0){\int }_{x_1}^{x_2}\frac{B(t)}{t^{s-s_0+1}}dt|\\ \\ & \le M{x}_2^{{\sigma }_0-\sigma }+M{x}_1^{{\sigma }_0-\sigma }+|s-s_0|M{\int }_{x_1}^{x_2}{t}^{{\sigma }_0-\sigma -1}dt\end{array}$$

Because $x_1<x_2$, we can say $M{x}_2^{{\sigma }_0-\sigma }+M{x}_1^{{\sigma }_0-\sigma }<2M{x}_1^{{\sigma }_0-\sigma }for\sigma >{\sigma }_0$. Despite appearances, evaluating the integral is easy.

$$\begin{array}{rl}\left|\sum _{x_1<n\le x_2}\frac{a_n}{n^s}\right|& \le 2M{x}_1^{{\sigma }_0-\sigma }+|s-s_0|M\left(\frac{x_2^{{\sigma }_0-\sigma }-x_1^{{\sigma }_0-\sigma }}{{\sigma }_0-\sigma }\right)\\ \\ & \le 2M{x}_1^{{\sigma }_0-\sigma }(1+\frac{|s-s_0|}{\sigma -{\sigma }_0})\end{array}$$

The last step uses $|x_2^{{\sigma }_0-\sigma }-x_1^{{\sigma }_0-\sigma }|=x_1^{{\sigma }_0-\sigma }-x_2^{{\sigma }_0-\sigma }<x_1^{{\sigma }_0-\sigma }<2x_1^{{\sigma }_0-\sigma }$. 

The key point is that $\sum _{x_1<n\le x_2}{a}_n/n^s$ is bounded if $\sum _{n\le x}{a}_n/n^{s_0}$ is bounded, where $\sigma >{\sigma }_0$.

Let's see if we can push this result about boundedness to convergence. 

$$\left|\sum _{x_1<n\le x_2}\frac{a_n}{n^s}\right|\le 2M{x}_1^{{\sigma }_0-\sigma }(1+\frac{|s-s_0|}{\sigma -{\sigma }_0})=K{x}_1^{{\sigma }_0-\sigma }$$

Here $K$ doesn't depend on $x_1$. If we let $x_1\to \mathrm{\infty }$ then $K{x}_1^{{\sigma }_0-\sigma }\to 0$, which means the magnitude of the tail of the infinite sum $\sum a_n/n^s$ diminishes to zero, and so the series is not just bounded, it also converges. 

Let's summarise our results so far:

• If $\sum _{n\le x}{a}_n/n^{s_0}$ is bounded, the infinite sum $\sum a_n/n^s$ converges for $\sigma >{\sigma }_0$.

• With the special case of $s_0=0$, if $\sum _{n\le x}{a}_n$ is bounded, the infinite sum $\sum a_n/n^s$ converges for $\sigma >0$.

The special case is particularly useful as we can sometimes say whether a series converges for $\sigma >0$ just by looking at the coefficients $a_n$.

Following the same logic as for ${\sigma }_a$, it is clear there is an abscissa of convergence ${\sigma }_c$ where a Dirichlet series converges for $\sigma >{\sigma }_c$, and diverges for $\sigma <{\sigma }_c$.

Maximum Difference Between ${\sigma }_c$ And ${\sigma }_a$

We know that not all convergent series are absolutely convergent, so we can say ${\sigma }_a\ge {\sigma }_c$. We shouldn't have to increase $\sigma$ by too much before a conditionally convergent series converges absolutely.

If a series converges at $s_0$, the magnitude of terms is bounded. We can call this bound $C$.

$$\sum \left|\frac{a_n}{n^s}\right|=\sum |\frac{a_n}{n^{s_0}}\cdot \frac{1}{n^{s-s_0}}|\le C\sum \frac{1}{n^{\sigma -{\sigma }_0}}$$

We know series of the form $\sum n^{{\sigma }_0-\sigma }$ only converge for ${\sigma }_0-\sigma >1$, so we can say if $\sigma$ is larger than ${\sigma }_c$ by at least 1, the series converges absolutely.

$$\boxed{{\displaystyle 0\le {\sigma }_a-{\sigma }_c\le 1}}$$

Example: Alternating Zeta Function

Let's apply our results to the alternating zeta function, also called the eta function.

$$\eta (s)=\sum \frac{(-1{)}^{n+1}}{n^s}=\frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\dots$$

At $s_0=0$ the partial sum $\sum _{n\le x}(-1{)}^{n+1}$ oscillates but is always bounded $\le 1$, and so $\sum (-1{)}^{n+1}/n^s$ converges for $\sigma >0$.