TITLE: An interesting series $\sum_{n=2}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(2k+3)^n-1}$
QUESTION [10 upvotes]: Here is an interesting series that unexpectedly (at least to me) evaluates to a nice value, that is 
$$\sum_{n=2}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(2k+3)^n-1}=1-\log(2)$$
As you can see, it's not one of the series we meet here daily, and I suppose the ways to approach
it might be a good lesson to learn. Any hint, clue for a good starting point is welcome.
Here is a supplementary question: Find the closed form of
$$\sum_{n=2}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(3k+4)^n-1}$$

REPLY [3 votes]: $$\frac{1}{(2k+3)^n-1}=\frac{1}{(2k+3)^n}+\frac{1}{(2k+3)^{2n}}+\frac{1}{(2k+3)^{3n}}+\ldots$$
and summing over $n$:
$$\sum_{n=2}^{+\infty}\frac{1}{(2k+3)^n-1}=\frac{\frac{1}{(2k+3)^2}}{1-\frac{1}{2k+3}}+\frac{\frac{1}{(2k+3)^4}}{1-\frac{1}{(2k+3)^2}}+\frac{\frac{1}{(2k+3)^6}}{1-\frac{1}{(2k+3)^3}}+\ldots$$
and summing over $k$:
$$\sum_{k=0}^{+\infty}\sum_{n=2}^{+\infty}\frac{1}{(2k+3)^n-1}=(1-\log 2)+\frac{1}{8}(10-\pi^2)+\ldots $$
so the first sum is greater than $1-\log 2$.