Swapping lim∑ For ∑lim

When we previously extended the Riemann Zeta function into the complex plane and visualised it, we observed that to the right, as $\sigma \to +\mathrm{\infty }$, the magnitude $|\zeta (s)|$ appears to approach 1.

The video for this topic is here [youtube], and slides here [pdf].

Let's consider what happens to the Riemann Zeta function $\zeta (s)as\sigma \to +\mathrm{\infty }$.

$$\underset{\sigma \to \mathrm{\infty }}{lim}\sum _n\frac{1}{n^s}=\underset{\sigma \to \mathrm{\infty }}{lim}(\frac{1}{1^s}+\frac{1}{2^s}+\dots )$$

It's tempting to look at each term and notice that $|n^{-s}|=n^{-\sigma }\to 0$ as $\sigma \to \mathrm{\infty }$ for all $n$ except $n=1$, then conclude $\zeta (s)\to 1$ as $\sigma \to \mathrm{\infty }$. In effect, we've taken the limit inside the sum.

$$\sum _n\underset{\sigma \to \mathrm{\infty }}{lim}\frac{1}{n^s}=\underset{\sigma \to \mathrm{\infty }}{lim}\left(\frac{1}{1^s}\right)+\underset{\sigma \to \mathrm{\infty }}{lim}\left(\frac{1}{2^s}\right)+\dots$$

However, the limit of an infinite sum is not always the sum of the limits. Tannery's theorem (also here, and here) tells us when we can swap sum and limit operators.

Tannery's Theorem

The theorem has three requirements

• An infinite sum $S_j=\sum _k{f}_k(j)$ that converges

• The limit $\underset{j\to \mathrm{\infty }}{lim}{f}_k(j)=f_k$ exists

• An $M_k\ge |f_k(j)|$ independent of $j$, where $\sum _k{M}_k$ converges

If the requirements are met, we can take the limit inside the sum.

$$\underset{j\to \mathrm{\infty }}{lim}\sum _k{f}_k(j)=\sum _k\underset{j\to \mathrm{\infty }}{lim}{f}_k(j)$$

Proof

Let's first show the sum of the limit actually exists.

By definition, $|f_k(j)|\le M_k$, and $\sum _k{M}_k$ converges. Taking $j\to \mathrm{\infty }$ gives us $\left|f_k\right|\le M_k$, and so $\sum _k\left|f_k\right|$ converges, which in turn means $\sum _k{f}_k$ converges absolutely. This quantity is the sum of limits.

Now let's show the limit of the sum is the sum of the limits.

Since $\sum _k{M}_k$ converges there must be an $N$ so that $\sum _{k=N}{M}_k<ϵ$, where $ϵ$ is as small as we require.

We can therefore say,

$$|\sum _{k=N}{f}_k(j)|\le \sum _{k=N}|f_k(j)|\le \sum _{k=N}{M}_k<ϵ$$

The following is the case when $j\to \mathrm{\infty }$.

$$\left|\sum _{k=N}{f}_k\right|\le \sum _{k=N}\left|f_k\right|\le \sum _{k=N}{M}_k<ϵ$$

Let's now consider the absolute difference between $\sum _k{f}_k(j)$ and $\sum _k{f}_k$. Although the following looks complicated, it is simply splitting the sums over $[0,\mathrm{\infty }]$ into sums over $[0,N-1]$ and $[N,\mathrm{\infty }]$.

$$\begin{array}{rl}|\sum _k{f}_k(j)-\sum _k{f}_k|& =|\sum _k^{N-1}{f}_k(j)+\sum _{k=N}{f}_k(j)-\sum _k^{N-1}{f}_k-\sum _{k=N}{f}_k|\\ \\ & \le |\sum _{k=N}{f}_k(j)|+\left|\sum _{k=N}{f}_k\right|+|\sum _k^{N-1}{f}_k(j)-\sum _k^{N-1}{f}_k|\\ \\ & <2ϵ+\left|\sum _k^{N-1}(f_k(j)-f_k)\right|\end{array}$$

As $j\to \mathrm{\infty }$, the finite sum $\sum _k^{N-1}(f_k(j)-f_k)\to 0$, which leaves a simpler inequality.

$$\underset{j\to \mathrm{\infty }}{lim}|\sum _k{f}_k(j)-\sum _k{f}_k|<2ϵ$$

Because $ϵ$ can be as small as we require, we have $\underset{j\to \mathrm{\infty }}{lim}\sum _k{f}_k(j)=\sum _k{f}_k$, which proves the theorem.

$$\underset{j\to \mathrm{\infty }}{lim}\sum _k{f}_k(j)=\sum _k{f}_k=\sum _k\underset{j\to \mathrm{\infty }}{lim}{f}_k(j)$$

Application To $\zeta (s)$

Let's apply Tannery's Theorem to $\zeta (s)$. Here $f_k(j)$ is written as $f_n(s)=1/n^s$.

We start with the convergent infinite sum.

$$\zeta (s)=\sum _n\frac{1}{n^s}\text{ converges for }\sigma >1$$

We confirm $f_n(s)$ exists when $\sigma \to \mathrm{\infty }$.

$$\underset{\sigma \to \mathrm{\infty }}{lim}\frac{1}{n^s}=f_n=\{\begin{array}{ll}1& n=1\\ 0& n>1\end{array}$$

We also find an $M_n\ge |f_n(s)|$ independent of $\sigma$.

$$\left|\frac{1}{n^s}\right|=\frac{1}{n^{\sigma }}\le M_n=\frac{1}{n^{\alpha }}$$

Here $1<\alpha \le \sigma$. The sum $\sum _n{M}_n$ converges because $\alpha >1$.

The criteria have been met, so we can legitimately move the limit inside the sum.

$$\underset{\sigma \to \mathrm{\infty }}{lim}\sum _n\frac{1}{n^s}=\sum _n\underset{\sigma \to \mathrm{\infty }}{lim}\frac{1}{n^s}=1+0+0+\dots$$

So $\zeta (s)\to 1$, as $\sigma \to +\mathrm{\infty }$. 

The Riemann Zeta Function is Almost Symmetric

This blog is a quick note on the symmetry of the Riemann Zeta function.

The video for this blog is online [youtube], and the slides here [pdf].

The plots of the magnitude $|\zeta (s)|$ that we rendered previously suggest the function is symmetric about the real axis. This isn't quite true. 

Remembering the complex conjugate $\bar{s}$ is a reflection of s in the real axis, for example $\overline{3+2i}=3-2i$, let's look again at the terms in the series $\zeta (s)=\sum 1/n^s$.

$$n^{-\bar{s}}=e^{-\bar{s}\ln (n)}=\overline{e^{-sln(n)}}=\overline{n^{-s}}$$

This means $\zeta (\bar{s})$ is the complex conjugate of $\zeta (s)$. 

So, although the magnitude of $\zeta (s)$ is mirrored above and below the real axis, the sign of the imaginary part is inverted.

Let's also consider the recently developed series $\zeta (s)=(1-2^{1-s}{)}^{-1}\eta (s)$. 

Following the same logic, we can say that $\eta (\bar{s})$ is the complex conjugate of $\eta (s)$, so this part has inverted phase above and below the real axis. 

We can say something similar for the other factor because $1/\bar{z}=\overline{1/z}$.

$$(1-2^{1-\bar{s}}{)}^{-1}=\overline{{(1-2^{1-s})}^{-1}}$$

Therefore we have:

$$\begin{array}{rl}\zeta (\bar{s})& =(1-2^{1-\bar{s}}{)}^{-1}\cdot \eta (\bar{s})\\ \\ & =\overline{(1-2^{1-s}{)}^{-1}}\cdot \overline{\eta (s)}\\ \\ & =\overline{\zeta (s)}\end{array}$$

This gives us the same conclusions that the magnitude is mirrored in the real axis, but the phase is inverted.

The plot below shows the phase of $\zeta (s)$ coloured to illustrate this almost-symmetry. 

ζ(s) Has Only One Pole In σ>0

Here we show that the Riemann Zeta function has only one pole in the domain $\sigma >0$. It is based on a suggestion by user @leoli1 on math.stackexchange.

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

Previously

The Riemann Zeta function represented by the series $\zeta (s)=\sum 1/n^s$ converges for $\sigma >1$, and therefore has no poles in that domain.

In the last blog post developed a new series for $\zeta (s)$ based on the eta function $\eta (s)$.

$$\zeta (s)=\frac{1}{1-2^{1-s}}\eta (s)$$

Because $\eta (s)$ converges for $\sigma >0$ (Dirichlet series), any divergence must come from the factor $(1-2^{1-s}{)}^{-1}$. 

The denominator $(1-2^{1-s})$ is zero at $s=1+2\pi ia/\ln (2)$ for integer $a$, so the factor $(1-2^{1-s}{)}^{-1}$ diverges at all these points. 

Visualising $\zeta (s)$ suggested it had only one pole at $s=1+0i$. If true, this would require $\eta (s)$ to have zeros at $s=1+2\pi ia/\ln (2)$ for integers $a\ne 0$ to cancel out the other poles from $(1-2^{1-s}{)}^{-1}$. 

To prove this directly isn't easy, but there is a nice indirect path.

Yet Another Series for $\zeta (s)$

We start with a specially constructed Dirichlet series.

$$X(s)=\frac{1}{1^s}+\frac{1}{2^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}-\frac{2}{6^s}+\dots$$

The pattern can be exploited to find yet another series for $\zeta (s)$.

$$\begin{array}{rl}\zeta (s)-X(s)& =\frac{3}{3^s}+\frac{3}{6^s}+\frac{3}{9^s}+\dots \\ \\ & =\frac{3}{3^s}\zeta (s)\\ \\ \zeta (s)& =\frac{1}{1-3^{1-s}}X(s)\end{array}$$

Comparing Potential Poles

The Dirichlet series $X(s)$ converges for $\sigma >0$ (Dirichlet series), so any divergence must come from the factor $(1-3^{1-s}{)}^{-1}$. The denominator $(1-3^{1-s})$ is zero when $s=1+2\pi ib/\ln (3)$ for integer $b$.

We can equate the two expressions for where the poles of $\zeta (s)$ could be.

$$\begin{array}{rl}1+\frac{2\pi ia}{\ln (2)}& =1+\frac{2\pi ib}{\ln (3)}\\ \\ \frac{a}{b}& =\frac{\ln (2)}{\ln (3)}\end{array}$$

There are no non-zero integers $a$ and $b$ which satisfy this because $\ln (2)/\ln (3)$ is irrational.

This leaves us with $s=1+0i$ as the only pole for $\zeta (s)$ in the domain $\sigma >0$.