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.
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.