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 $\sigma \to +\mathrm{\infty }$.