I ran across a cool series I have been trying to chip away at.
$\sum_{k=1}^{\infty}\frac{\zeta(2k+1)-1}{k+2}=\frac{-\gamma}{2}-6\ln(A)+\ln(2)+\frac{7}{6}\approx 0.0786\ldots$
where A = the Glaisher-Kinkelin constant. Numerically, it is approx. $1.282427\ldots$
I began by writing zeta as a sum and switching the summation order
$\sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(k+2)n^{2k+1}}$
The first sum is the series for $-n^3\ln(1-\frac{1}{n^2})-n-\frac{1}{2n}$
So, we have $-\sum_{n=2}^\infty \left[\ln(1-\frac{1}{n^2})+n+\frac{1}{2n}\right]$
This series numerically checks out, so I am onto something. At first glance the series looks like it should diverge, but it does converge.
Another idea I had was to write out the series of the series:
$1/3(1/2)^{3}+1/4(1/2)^{5}+1/5(1/2)^{7}+\cdots +1/3(1/3)^{3}+1/4(1/3)^{5}+1/5(1/3)^{7}+\cdots +1/3(1/4)^{3}+1/4(1/4)^{5}+1/5(1/4)^{7}+\cdots$
and so on.
This can be written as $1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots +1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots + 1/3x^{3}+1/4x^{5}+1/5x^{7}+\cdots $
where $x=1/2,1/3,1/4,\ldots$
This leads to the series representation for:
$\frac{-\ln(1-x^2)}{x^3}-\frac{1}{x}-\frac{x}{2}$
Since $x$ is of the form $1/n$, we end up with the same series as before.
Now, my quandary. How to finish?. Where in the world does that Glaisher-Kinkelin constant come in, and how can that nice closed from be obtained?. Whether from the series I have above or some other means. As usual, it is probably something I should be seeing but don't at the moment.
The GK constant has a closed form of $e^{\frac{1}{12}-\zeta^{'}(-1)}$.
Which means an equivalent closed form would be $\frac{-\gamma}{2}+\ln(2)+6\zeta^{'}(-1)+\frac{2}{3}$
Thanks all.