Although $\eta(1)$ is known to be $\ln(2)$, I have not seen an analytically calculated value for $\eta(\frac{1}{2});$

$$\eta\left(\frac{1}{2}\right) = \sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{\sqrt{n}}$$

A web calculator gives the value to be 0.6, which seems to be right.

up vote 11 down vote accepted

Isn't just $$\eta\left(\frac{1}{2}\right)=\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{\sqrt{n}}=\left(1-\sqrt{2}\right) \zeta \left(\frac{1}{2}\right)\approx 0.6048986434$$


Remember the general relation $$\eta\left(s\right)=\left(1-2^{1-s}\right) \zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 \leq s \leq 1$, you could use $$\eta\left(s\right)=\frac 12+\left( \log (2)-\frac{1}{2}\right)\, s^{0.895}$$

A careful computation shows that the numerical value is $$0.6048986434216303702472...$$ which is not $0.6$. One should be aware that the above series converge really slowly.

As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $\zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $\eta(1/2)$".

EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$\gamma/2 + \pi/4 - (1/2 + \sqrt{2})\log(2) + \log(\pi)/2,$$where $\gamma$ is the Euler-Mascheroni constant, is $\eta'(1/2)/\eta(1/2)$ not $\eta(1/2)$.

  • You are very correct ! We start a no-end loop. By the way $\to +1$ – Claude Leibovici Nov 13 at 10:51
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    It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though. – leftaroundabout Nov 13 at 16:01
  • @leftaroundabout Do you have a reference, I searched, but I didn't find it. – Josué Tonelli-Cueto Nov 13 at 18:47
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    @user3059799 Corrected, editing from the phone is hard – Josué Tonelli-Cueto Nov 13 at 20:11
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    The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $\zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise. – Mike Miller Nov 14 at 1:18

The numerical value of 0.604898... is provided in .

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