an infinite series involving odd zeta 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. 
 A: We have
$$
\begin{eqnarray*}
&&\sum_{n=2}^\infty\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\
&=&\lim_{N\to\infty}\sum_{n=2}^N\left( -n^3\log(1-\frac{1}{n^2})-n-\frac{1}{2n}\right)\\
&=&\lim_{N\to\infty}\left(\sum_{n=2}^N -n^3\log(1-\frac{1}{n^2})-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}+O\left(\frac{1}{N}\right)\right)\right)\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N (2n^3\log(n)-n^3\log(n+1)-n^3\log(n-1))-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]
\end{eqnarray*}
$$
In the sum on the last line, we may gather together the coefficients of each logarithm (terms at the boundary of the sum are a little funny), giving
$$
\begin{eqnarray*}
&&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)+(N^3+3N^2+3N+1)\log(N)-N^3\log(N+1)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\log\left(1+\frac{1}{N}\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\log(2)-N^3\left(\frac{1}{N}-\frac{1}{2N^2}+\frac{1}{3N^3}+O\left(\frac{1}{N^4}\right)\right)+(3N^2+3N+1)\log(N)-\left(\frac{N^2+N-2}{2}\right)-\left(\frac{\log(N)}{2}-\frac{1}{2}+\frac{\gamma}{2}\right)]\\
&=&\lim_{N\to\infty}[\sum_{n=2}^N(-6n\log(n))+\left(3N^2+3N+\frac{1}{2}\right)\log(N)+\left(\frac{-3}{2}N^2+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\right)]\\
&=&-6\log\left(\lim_{N\to\infty}\left(\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}}\right)\right)+\frac{7}{6}-\frac{\gamma}{2}+\log(2)\\
&=&-6\log(A)+\frac{7}{6}-\frac{\gamma}{2}+\log(2)
\end{eqnarray*}
$$
Here, I am taking
$$
A=\lim_{N\to\infty}\frac{\prod_{n=1}^N n^n}{N^{N^2/2+N/2+1/12}e^{-N^2/4}}
$$
as the definition of the Glaisher-Kinkelin constant.
