Let $X \sim \mathcal{B}(Nm,p)$ with parameters $p$ (probability of success) and $Nm$ (the number of trials).
In the article I'm reading, it is said that :
$\mathbb{P}(X < N) \leq N(1-p)^m$
Is there an easy/clean way to get this upper bound ?
Writing $\mathbb{P}(X < N)$ as a sum of $N$ terms of the form ${{Nm}\choose{i}} p^i (1-p)^{Nm-i}$, it seems true (numerically at least) that $(1-p)^m$ is an upper bound for each of the summands, but this is kind of tedious...
Chop up your sequence of $Nm$ trials into $N$ disjoint segments of $m$ consecutive trials. If it is true that $X<N$, then at least one of these segments consists of all failures. It follows that $$ P(X<N)\le P(\bigcup_{i=1}^N \{\text{no successes occur in segment $i$}\})\le\sum_{i=1}^NP(\text{no successes occur in segment $i$})$$ The RHS simplifies to $\sum_{i=1}^N (1-p)^m=N(1-p)^m$.