An interesting series $\sum_{n=2}^{\infty} \sum_{k=0}^{\infty} \frac{1}{(2k+3)^n-1}$
Solution 1:
$$\frac{1}{(2k+3)^n-1}=\frac{1}{(2k+3)^n}+\frac{1}{(2k+3)^{2n}}+\frac{1}{(2k+3)^{3n}}+\ldots$$ 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 $1-\log 2$.