Given a value $p\in[0,1]$ (a probability of occurrence), I would like to bound the following expression:
$$ s\frac{1-(1-p)^{k+1}}{p(k+1)} + (1-s)\frac{1-(1-p)^{k}}{pk},\ \ \ \text{where $k=\lfloor 1/p \rfloor$ and $s=1/p-k$}.$$
I would like to prove that the latter is bigger than $1-e^{-1}$.
I already know that in each interval where $k$ is constant ($p\in (\frac{1}{k+1},\frac{1}{k}]$), when $p$ approaches to $1/(k+1)$, the expression is bounded by $1-e^{-1}$ from below. Therefore, if I could prove that in such interval the function is increasing in $p$, I would be done.
I've struggling a while trying to prove this (and I have also tried to prove the inequality directly), but I have failed so far.
I think this kind of expressions are not that ''unnatural'' to come out, so maybe somebody else has already studied them. However, I don't know where to look. All this work is in the context of approximation theory and optimization; I have not found similar expressions though.
Any ideas on how I could keep the work or where I should look for similar results? Thanks!