Erroneous numerical approximations of $\zeta\left(\frac{1}{2}\right)$?

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.

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

$$ \Gamma(s/2) \pi^{-s/2} \zeta(s) = \Gamma\left(\frac{1-s}{2}\right) \pi^{- \frac{1-s}{2}}\zeta(1-s). $$

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")