Negative Log Probability

For $\lfloor-\ln x\rfloor$ to equal an even number $2k$ (where k is an integer), then $2k+1>-\ln x\geq2k\\ -2k-1<\ln x\leq-2k\\ e^{-2k-1}<x\leq e^{-2k}$ Now, $x$ must be between 0 and 1. When $k=0,e^{-1}<x\leq 1$ and as $k\to\infty,e^{-2k-1}\to0,e^{-2k}\to0$ . Since all functions here are continuous and monotonically increasing or decreasing, we can conclude that all possible values of $x$ between 0 and 1 are accounted for by all non-negative integer values of $k$ (i.e. $k=0,1,2,\dots$ ).

The size of the region for a certain value of $k$ is the upper bound minus the lower bound, i.e. $e^{-2k}-e^{-2k-1}$ . Therefore the sum of all the regions where $\lfloor-\ln x\rfloor$ is an even number is equal to $\sum_{k=0}^\infty e^{-2k}-e^{-2k-1}\\ =e^0-e^{-1}+e^{-2}-e^{-3}+e^{-4}-e^{-5}\dots\\ =\sum_{n=0}^\infty (-1)^n e^{-n}\\ =\frac{1}{1-e^{-1}}\\ =\boxed{\frac{e}{e+1}}$