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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 $p=\frac{1}{2}$.

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    $\begingroup$ The standard Dirichlet series representation of the Riemann zeta function $\zeta(s)$ does not converge for $\mathrm{Re}(s) \leq 1$. $\endgroup$ – Gahawar Sep 26 '14 at 16:08
  • $\begingroup$ Did you try the obvious? $\endgroup$ – Did Sep 26 '14 at 16:11
  • $\begingroup$ I will try this. Thanks for pointing out. $\endgroup$ – max Sep 26 '14 at 17:30
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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 $x>0$. :-$)$

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  • $\begingroup$ For your first equation, why is it true for $x>1$, but you conclude a result for $x>0$? [I can't immediately see it, so even if you say something like "Equation x justifies it" would be appreciated :)] $\endgroup$ – Alex Nelson Sep 26 '14 at 16:59
  • $\begingroup$ @AlexNelson: For $x\le1$, the first expression diverges. Obviously, $\dfrac12<1$. However, due to the alternating series test, the expression for Dirichlet's $\eta$ function converges for all $x>0$, making the new expression for $\zeta(x)$ converge as well. $\endgroup$ – Lucian Sep 26 '14 at 17:08
  • $\begingroup$ @Lucian Thanks for this elaborated analysis. I will redo the equations by myself and I think, this would point towards the solution. $\endgroup$ – max Sep 26 '14 at 17:37

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