I came out with a geometric sum with a floor function in the upper limit. It looks somehow like this:
$$ \sum_{i=0}^{\lfloor 1/p \rfloor} (1-p)^i,\ \ 0<p<1.$$
I would like to study some properties about it. More specifically, what I have is a convex combination of the form
$$s(1-(1-p)^{k}) + (1-s)(1-(1-p)^{k+1}),\ \ \text{where $k=\lfloor 1/p \rfloor$ and $s=1/p-k$,} $$
and I would like to bound (from below) that expression.
Do you know where I could look? Has anyone worked on this or something similar? I've been stuck trying to get somewhere, and I suppose that - since the expression looks ''familiar'' - someone could have already crossed with it but I don't know where could be the best place to start.
A friend helped me with this.
A way is to compute the derivative of that expression for $\frac{1}{k+1}\leq p \leq \frac{1}{k}$, which is equal to $k(1-p)^{k-1}((k+1)p-1)$, which is positive.
Then, for $p$ approaching to $1/(k+1)$, you always get a lower bound of $1-e^{-1}$.