Erroneous numerical approximations of $\zeta\left(\frac{1}{2}\right)$? By definition of the Riemann Zeta Function, $$\zeta\left(\frac{1}{2}\right) = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}.$$ Since $\forall n \geq 1 : \frac{1}{\sqrt{n}} \geq \frac{1}{n}$, we have that for all $N \geq 1$, $$\sum_{n=1}^N\frac{1}{\sqrt{n}} \geq \sum_{n=1}^N \frac{1}{n},$$ but it is well known that $$\lim_{N\rightarrow\infty}\sum_{n=1}^N\frac{1}{n}=\infty,$$ so $\zeta\left(\frac{1}{2}\right)$ diverges by the comparison test.
In other words, $\zeta\left(\frac{1}{2}\right)$ should equal positive infinity, correct?  If so, why do Maple, Mathematica, and Matlab all return a value of around $-1.4604$ when asked to numerically approximate this value?  For example, see here.
 A: We have such a thing even with geometric series (which has many things easier than the Dirichlet/zeta-series).
Consider $s_2=1+2+4+8+...+2^k+...$ and $s_3=3+9+27+81+...+3^k+...$ .
Although each term of $s_3$ is bigger that that of the same index k in $s_2$ (and both sums are diverging), the sum $s_2=-1 (={1 \over 1-2}) $ is bigger that $s_3 = -{3 \over 2} (={3 \over 1-3})$ (You may look for "geometric series/analytic continuation")
A: The analytic continuation of $\zeta(s)$ for $Re(s)>0$ is given by
$$\zeta(s) = \frac{1}{1-2^{1-s} } \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.$$
So this is why WolframAlpha gives you
$$\zeta \left( \frac12 \right) = -(1+\sqrt{2}) \sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt{n}}.$$
See here for more details.
A: To supplement Derek's answer, the Riemann zeta function $\zeta(s) = \sum_{n} n^{-s}$ was originally only considered for $s = \sigma + i t$, where $\sigma > 1$.  Note that when $\sigma > 1$, the zeta function always converges absolutely.  However, Riemann showed that the definition of the $\zeta$ function could be extended via analytic continuation to the whole complex plane (with a pole at $s = 1$).  There, it satisfies the functional equation  
$$
\Gamma(s/2) \pi^{-s/2} \zeta(s) = \Gamma\left(\frac{1-s}{2}\right) \pi^{- \frac{1-s}{2}}\zeta(1-s).
$$
A: To the formula:
$\zeta(s)=\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}$
is written in small print: $1<\Re(s)$
That's a classical trap. Here my solution:
According to the russian Wikipedia :
$\zeta(s)=\displaystyle\lim_{N \to \infty}(\displaystyle\sum_{n=1}^{N} \frac{1}{n^s} -\frac{N^{1-s}}{1-s})$ ;for $\Re(s)>0$ and $\Re(s)\ne 1$.
By putting s=1/2 , we get:
$\zeta(1/2)=\displaystyle\lim_{N \to \infty}(\displaystyle\sum_{n=1}^{N} \frac{1}{\sqrt{n}} -2\sqrt{N})$
This series converges very slowly but converge.
