Ytukyg

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

$zeta(s)=sum_{n} n^{-s}$ 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 $s>1$:

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

The right side is defined for any $sin(0,1]$, 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 $sin(0,1)$:

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

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.

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

[**] We have a general theorem:

Given a sequence of complex numbers ${a_i}$ such that $a_ito 0$ 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 $(1-x)^s=1-xs+o(x)$:

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

$$zeta(s) = sum_{n=1}^{infty} dfrac1{n^s}$$
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.

According to Gradshteyn-Ryzhik (1980) 9.513.1, we have
$$
zeta(s)=frac{1}{(1-2^{1-s})Gamma(s)}
int_0^inftyfrac{t^{s-1} text{d}t}{e^t+1},
qquad (text{Re} s>0)
$$
which is rewritten by partial integration (because $text{Re} s>0$) as
$$
zeta(s)=
frac{1}{(1-2^{1-s})Gamma(s+1)}
int_0^infty frac{t^s e^t text{d}t}{(e^t+1)^2}.
$$
By changing the integration variable $t=2x$, we obtain
$$
zeta(s)=frac{2^{s-1}}{(1-2^{1-s})Gamma(s+1)}
int_0^inftyfrac{x^s text{d}x}{text{cosh}^2 x},
qquad (text{Re} s>0)
$$
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 $zeta(1/2)$:
$$
zeta(1/2)=-(sqrt{2}+1)sqrt{frac{2}{pi}}
int_0^inftyfrac{sqrt{x} text{d}x}{text{cosh}^2 x},
$$
which can be numerically integrated accurately such as the double exponential method of Mori-Takahashi. The result is
$$
zeta(1/2)=-1.460354508809586cdots,
$$
which is our final result.

No. The formula $zeta(s)=sum_{n=1}^inftyfrac1{n^s}$ is only valid for $Re s>1$. The function is defined for other values of $s$ by analytic continuation, using the functional equation.

According to the russian Wikipedia :

$zeta(s)=displaystylelim_{N to infty}(displaystylesum_{n=1}^{N} frac{1}{n^s} -frac{N^{1-s}}{1-s})$ ;for $Re(s)>0$ and $Re(s)ne 1$.

By putting s=1/2 , we get:

$zeta(1/2)=displaystylelim_{N to infty}(displaystylesum_{n=1}^{N} frac{1}{sqrt{n}} -2sqrt{N})$

This series converges very slowly

(I got for N=14000 the result $zeta(1/2)approx -1.45613 $ which still away from $approx -1.46035$)