I would like to calculate the sum
$$ \sum_{x = 1}^{\infty}\frac{1}{x}\, \log\left(\frac{kx}{\left\lfloor kx \right\rfloor}_{o}\right) $$
where $k = \sqrt{\, 2\,} + 1$, $x$ is an odd integer and $\left\lfloor z \right\rfloor_o$ indicates the greatest odd integer $\leq z$. Considering that the odd greatest integer function satisfies $\left\lfloor z \right\rfloor_{o} = 2\left\lfloor\frac{z + 1}{2}\right\rfloor - 1$, and setting $x = 2n - 1$ to obtain a summation over all positive integers, the sum can also be written as
$$ \sum_{n = 1}^{\infty}\frac{1}{2n - 1}\, \log\left(\frac{k\left[2n - 1\right]} {2\left\lfloor kn - k/2 + 1/2\right\rfloor - 1}\right) $$
The numerical value of the sum converges to $0.952842\ldots$ ( these first six decimal digits are stable after summing the first $10^{8}$ terms ). I tried several approaches to determine an explicit expression for this limit, including transformations of floor function by Fourier series or Laplace transforms, but I was not able to get it. Even if a closed expression could maybe not exist, I would be very interested in obtaining a simpler expression for this infinite sum.