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

9You're confusing between the summation $\sum_{n>0}\frac1{n^s}$ and the $\zeta$regularization of the sum. – Asaf Karagila Jan 15 '16 at 14:32

4$\zeta(s)$ is actually defined for _every_ complex number except for $s=1$. The series for $\zeta$ doesn't always converge, but the series isn't exactly the definition of $\zeta$. – Akiva Weinberger Jan 15 '16 at 14:33

1There is a formula for this number at https://oeis.org/A059750 The user anon here at math overflow knows how to explain it. I wrote the formula in the chat room and anon then referred to this answer of his:http://math.stackexchange.com/a/113837/8530 – Mats Granvik Jan 15 '16 at 14:35

3By the way, $\zeta(1/2)=(\sqrt2+1)\sum_{k=1}^\infty\frac{(1)^k}{\sqrt k}$, if you want some converging sequence that defines it. – Akiva Weinberger Jan 15 '16 at 14:40
5 Answers
$\zeta(s)=\sum_{n} n^{s}$ when that makes sense. But $\zeta$ has an "analytic continuation" to much of the rest of the complex plane.
Consider the equality:
$$\frac{1}{1z} =\sum_{n=0}^{\infty} z^n$$
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 $\frac{1}{1z}$ 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 $s>1$:
$$\left(1\frac{1}{2^{s1}}\right)\sum_{n} \frac{1}{n^s} =\sum_{n}\frac{1}{n^s}  2\sum_{n} \frac{1}{(2n)^s}=\sum_{n} \frac{(1)^{n1}}{n^s}$$
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 $s\in(0,1)$:
$$\zeta(s) = \frac{1}{1\frac{1}{2^{s1}}} \sum_n \frac{(1)^{n1}}{n^s}$$
This is equal to our original definition when $\mathrm{Re }\,s>1$.
So $$\zeta(1/2) = \frac{1}{1\sqrt{2}}\sum_{n} \frac{(1)^{n1}}{\sqrt{n}}$$
Computing for $M=200,000,000$ terms of this sum I get:
$$\frac{1}{1\sqrt{2}}\sum_{n=1}^M \frac{(1)^{n1}}{\sqrt{n}}\approx 1.460269$$
This series converges very slowly.
[*] Just for clarity: If you have a pathconnected 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 $$\sum_{i=1}^{\infty} (a_{2i1}+a_{2i})$$ converges, then so does $\sum_{i=0}^{\infty} a_i$.
In this case, let $a_i=\frac{(1)^{i1}}{i^s}$. Then we'll use that $(1x)^s=1xs+o(x)$:
$$\begin{align}a_{2i1}+a_{2i} &=\frac{(2i)^s(2i1)^s}{(2i(2i1))^s}\\ &=\frac{1\left(1\frac{1}{2i}\right)^s}{(2i1)^s}\\ &=\frac{\frac{s}{2i} + o\left(\frac{1}{i}\right)}{(2i1)^s}\\ &=O\left(\frac{1}{i^{s+1}}\right) \end{align}$$
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"any function has at most one analytic continuations to the regions where it is not always defined" ... perhaps you need to reconsider... – GEdgar Jan 15 '16 at 19:05

Yeah, meant "already." And technically, it isn't true for disconnected regions, I was trying to not get into the nittygritty about analytic continuations, but may have oversimplified. – Thomas Andrews Jan 15 '16 at 19:27

In a certain sense, it can also fail for nonsimply connected regions. Such as extending $z^{1/2}$ from the right halfplane to the punctured plane. So, in particular, some work must be done when extending $\zeta(z)$ to make sure it is unique. – GEdgar Jan 15 '16 at 21:28

No, there is no way to analytically extend $z^{1/2}$ to the punctured plane. Which agrees with my statement. @GEdgar – Thomas Andrews Jan 15 '16 at 21:33

I recognize what you are saying  that the uniqueness is up to the region in which you want to define. I was not trying to be formal  the OP might or might not have the background for a detailed description of what that means. – Thomas Andrews Jan 15 '16 at 21:44
According to GradshteynRyzhik (1980) 9.513.1, we have $$ \zeta(s)=\frac{1}{(12^{1s})\Gamma(s)} \int_0^\infty\frac{t^{s1}\ \text{d}t}{e^t+1}, \qquad (\text{Re}\ s>0) $$ which is rewritten by partial integration (because $\text{Re}\ s>0$) as $$ \zeta(s)= \frac{1}{(12^{1s})\Gamma(s+1)} \int_0^\infty \frac{t^s\ e^t\ \text{d}t}{(e^t+1)^2}. $$ By changing the integration variable $t=2x$, we obtain $$ \zeta(s)=\frac{2^{s1}}{(12^{1s})\Gamma(s+1)} \int_0^\infty\frac{x^s\ \text{d}x}{\text{cosh}^2 x}, \qquad (\text{Re}\ s>0) $$ where the convergence condition may be extended to $\text{Re}\ s>1$, although we need not such extention now.
By putting $s=1/2$, we have an integral representation of $\zeta(1/2)$: $$ \zeta(1/2)=(\sqrt{2}+1)\sqrt{\frac{2}{\pi}} \int_0^\infty\frac{\sqrt{x}\ \text{d}x}{\text{cosh}^2\ x}, $$ which can be numerically integrated accurately such as the double exponential method of MoriTakahashi. The result is $$ \zeta(1/2)=1.460354508809586\cdots, $$ which is our final result.
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$$\zeta(s) = \sum_{n=1}^{\infty} \dfrac1{n^s}$$ 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.
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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.
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According to the russian Wikipedia :
$\zeta(s)=\displaystyle\lim_{N \to \infty}(\displaystyle\sum_{n=1}^{N} \frac{1}{n^s} \frac{N^{1s}}{1s})$ ;for $\Re(s)>0$ and $\Re(s)\ne 1$.
By putting s=1/2 , we get:
$\zeta(1/2)=\displaystyle\lim_{N \to \infty}(\displaystyle\sum_{n=1}^{N} \frac{1}{\sqrt{n}} 2\sqrt{N})$
This series converges very slowly
(I got for N=14000 the result $\zeta(1/2)\approx 1.45613 $ which still away from $\approx 1.46035$)
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