Why $\zeta (1/2)=-1.4603545088...$?

Question

I saw $\zeta (1/2)=-1.4603545088...$ in this link. But how can that be? Isn't $\zeta (1/2)$ divergent since $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..$ ?

Answer 1

$$\zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s}$$ only when $\textbf{Real part}\mathbf{(s) > 1}$. For the rest of $s$, in the complex plane, it is defined as the analytic continuation of the above function.

Answer 2

$\zeta(s)=\sum_{n} n^{-s}$ when that makes sense. But $\zeta$ has an "analytic continuation" to much of the rest of the complex plane.

Consider the equality:

$$\frac{1}{1-z} =\sum_{n=0}^{\infty} z^n$$

The right side only converges and equals the left side when $|z|<1$.

But complex analysis has this awesome feature - that any function has at most one analytic continuations to the regions where it is not already defined.[*] So $\frac{1}{1-z}$ is the only analytic continuation of the right side for complex $z$.

The same is true for the $\zeta$ function. We first define it for $s$ where the real part is greater than one. And then we find a way to continue that function for other values of $s$.

The heart of the extension of $\zeta$, at least to $1/2$, is that, for $s>1$:

$$\left(1-\frac{1}{2^{s-1}}\right)\sum_{n} \frac{1}{n^s} =\sum_{n}\frac{1}{n^s} - 2\sum_{n} \frac{1}{(2n)^s}=\sum_{n} \frac{(-1)^{n-1}}{n^s}$$

The right side is defined for any $s\in(0,1]$, since it is the sum of an alternating decreasing sequence. (It converges for other $s$ with $\mathrm{Re}\,s >0$, but it isn't 100% obvious looking at it that this is true.[**])

This lets us extend $\zeta(s)$ to $s\in(0,1)$:

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

This is equal to our original definition when $\mathrm{Re }\,s>1$.

So $$\zeta(1/2) = \frac{1}{1-\sqrt{2}}\sum_{n} \frac{(-1)^{n-1}}{\sqrt{n}}$$

Computing for $M=200,000,000$ terms of this sum I get:

$$\frac{1}{1-\sqrt{2}}\sum_{n=1}^M \frac{(-1)^{n-1}}{\sqrt{n}}\approx -1.460269$$

This series converges very slowly.

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[*] Just for clarity: If you have a path-connected open subset $U\subseteq \mathbb C$ and $V\subseteq U$ so that $V$ contains an bounded infinite subset, and a function $f:V\to\mathbb C$, then there is at most one analytic $g:U\to \mathbb C$ so that $g(v)=f(v)$ for $v\in V$.

[**] We have a general theorem:

Given a sequence of complex numbers $\{a_i\}$ such that $a_i\to 0$ and $$\sum_{i=1}^{\infty} (a_{2i-1}+a_{2i})$$ converges, then so does $\sum_{i=0}^{\infty} a_i$.

In this case, let $a_i=\frac{(-1)^{i-1}}{i^s}$. Then we'll use that $(1-x)^s=1-xs+o(x)$:

$$\begin{align}a_{2i-1}+a_{2i} &=\frac{(2i)^s-(2i-1)^s}{(2i(2i-1))^s}\\ &=\frac{1-\left(1-\frac{1}{2i}\right)^s}{(2i-1)^s}\\ &=\frac{\frac{s}{2i} + o\left(\frac{1}{i}\right)}{(2i-1)^s}\\ &=O\left(\frac{1}{i^{s+1}}\right) \end{align}$$

Answer 3

No. The formula $\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$ is only valid for $\Re s>1$. The function is defined for other values of $s$ by analytic continuation, using the functional equation.