This is from Feller's Introduction to Probability Theory and Its Applications. In the context of Bernoulli trials, we define:
$b(k;n,p) = \binom{n}{k}p^kq^{n-k},$ $P\{S_n \ge r\} = \sum_{v=0}^{\infty}b(r+v;n,p).$
The latter being the probability of having at least $r$ successes. Now, supposing $r \gt np$ and knowing that
$\frac{b(k; n,p)}{b(k-1;n,p)}=\frac{(n-k+1)p}{kq}=1+\frac{(n+1)p-k}{kq},$
show that
$P\{S_n \ge r\} \le b(r;n,p)\frac{rq}{r-np}.$
According to Feller, it follows from the obvious fact that the terms of the series decrease faster than the terms of a geometric series with ratio $1-\frac{r-np}{rq}$. However, it's not obvious for me and I don't see how the upper bound follows.