Swapping $\lim \sum$ For $\sum \lim$

When we previously extended the Riemann Zeta function into the complex plane and visualised it, we observed that to the right, as $\sigma \rightarrow +\infty$, 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 $\zeta(s) as \sigma\rightarrow+\infty$.

$$\lim_{\sigma\rightarrow\infty}\sum_{n}\frac{1}{n^{s}}=\lim_{\sigma\rightarrow\infty}\left(\frac{1}{1^{s}}+\frac{1}{2^{s}}+\ldots\right)$$

It's tempting to look at each term and notice that $|n^{-s}|=n^{-\sigma}\rightarrow0$ as $\sigma\rightarrow\infty$ for all $n$ except $n=1$, then conclude $\zeta(s)\rightarrow1$ as $\sigma\rightarrow\infty$. In effect, we've taken the limit inside the sum.

$$\sum_{n}\lim_{\sigma\rightarrow\infty}\frac{1}{n^{s}}=\lim_{\sigma\rightarrow\infty}\left(\frac{1}{1^{s}}\right)+\lim_{\sigma\rightarrow\infty}\left(\frac{1}{2^{s}}\right)+\ldots$$

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 $\lim_{j\rightarrow\infty}f_{k}(j)=f_{k}$ exists

• An $M_{k}\geq\left|f_{k}(j)\right|$ independent of $j$, where $\sum_{k}M_{k}$ converges

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

$$\lim_{j\rightarrow\infty}\sum_{k}f_{k}(j)=\sum_{k}\lim_{j\rightarrow\infty}f_{k}(j)$$

Proof

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

By definition, $\left|f_{k}(j)\right|\leq M_{k}$, and $\sum_{k}M_{k}$ converges. Taking $j\rightarrow\infty$ gives us $\left|f_{k}\right|\leq M_{k}$, 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 $\sum_{k=N}M_{k}<\epsilon$, where $\epsilon$ is as small as we require.

We can therefore say,

$$\left|\sum_{k=N}f_{k}(j)\right|\leq\sum_{k=N}\left|f_{k}(j)\right|\leq\sum_{k=N}M_{k}<\epsilon$$

The following is the case when $j\rightarrow\infty$.

$$\left|\sum_{k=N}f_{k}\right|\leq\sum_{k=N}\left|f_{k}\right|\leq\sum_{k=N}M_{k}<\epsilon$$

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 $[0,\infty]$ into sums over $[0,N-1]$ and $[N,\infty]$.

$$\begin{align}\left|\sum_{k}f_{k}(j)-\sum_{k}f_{k}\right|&=\left|\sum_{k}^{N-1}f_{k}(j)+\sum_{k=N}f_{k}(j)-\sum_{k}^{N-1}f_{k}-\sum_{k=N}f_{k}\right| \\ \\&\leq\left|\sum_{k=N}f_{k}(j)\right|+\left|\sum_{k=N}f_{k}\right|+\left|\sum_{k}^{N-1}f_{k}(j)-\sum_{k}^{N-1}f_{k}\right| \\ \\ &<2\epsilon+\left|\sum_{k}^{N-1}\left(f_{k}(j)-f_{k}\right)\right|\end{align}$$

As $j\rightarrow\infty$, the finite sum $\sum_{k}^{N-1}\left(f_{k}(j)-f_{k}\right)\rightarrow0$, which leaves a simpler inequality.

$$\lim_{j\rightarrow\infty}\left|\sum_{k}f_{k}(j)-\sum_{k}f_{k}\right|<2\epsilon$$

Because $\epsilon$ can be as small as we require, we have $\lim_{j\rightarrow\infty}\sum_{k}f_{k}(j)=\sum_{k}f_{k}$, which proves the theorem.

$$\lim_{j\rightarrow\infty}\sum_{k}f_{k}(j)=\sum_{k}f_{k}=\sum_{k}\lim_{j\rightarrow\infty}f_{k}(j)$$

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 $\sigma\rightarrow\infty$.

$$\lim_{\sigma\rightarrow\infty}\frac{1}{n^{s}}=f_{n}=\begin{cases} 1 & n=1\\ 0 & n>1 \end{cases}$$

We also find an $M_{n}\geq\left|f_{n}(s)\right|$ independent of $\sigma$.

$$\left|\frac{1}{n^{s}}\right|=\frac{1}{n^{\sigma}}\leq M_{n}=\frac{1}{n^{\alpha}}$$

Here $1<\alpha\leq\sigma$. 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.

$$\lim_{\sigma\rightarrow\infty}\sum_{n}\frac{1}{n^{s}}=\sum_{n}\lim_{\sigma\rightarrow\infty}\frac{1}{n^{s}}=1+0+0+\ldots$$

So $\zeta(s)\rightarrow1$, as $\sigma\rightarrow+\infty$.