This may be a silly question, but I need to figure out how to evaluate the value of $\zeta(\frac{1}{2})$. In wikipedia, it says: $\zeta(1/2) \approx -1.4603545$. I am interested to know how this value is calculated. Should it not be a positive number? I am applying this to probability related calculations.
In fact my current problem is to evaluated the sum: $\sum_{j=1}^{m}\sum_{i=j}^{n}\frac{1}{i^p}$, where $n > 1$, $1 < m <n$, and $0 \leq p \leq 1$ is constant. I am assuming $n \to \infty$. I am specially interested for the case where $p=\frac{1}{2}$.
For $x>1$ we have $$\zeta(x)=\sum_1^\infty\frac1{n^x}=\sum_1^\infty\frac1{(2n)^x}+\sum_1^\infty\frac1{(2n-1)^x}=2^{-x}\zeta(x)+\sum_1^\infty\frac1{(2n-1)^x}$$ from where we deduce that $$\sum_1^\infty\frac1{(2n-1)^x}=(1-2^{-x})\zeta(x)$$ which then leads us to $$\eta(x)=-\sum_1^\infty\frac{(-1)^n}{n^x}=-\sum_1^\infty\frac{(-1)^{2n}}{(2n)^x}-\sum_1^\infty\frac{(-1)^{2n-1}}{(2n-1)^x}=-2^{-x}\zeta(x)+\sum_1^\infty\frac1{(2n-1)^x}$$ which, in its turn, allows us to write $$\eta(x)=(1-2^{1-x})\zeta(x)\iff\zeta(x)=\dfrac{\eta(x)}{1-2^{1-x}}$$ the advantage of which is that the latter expression for $\zeta(x)$ converges for all $x>0$. :-$)$