Upper bound on first $k$ terms of binomial expansion where the variable is probability If $p$ is some probability, meaning $0<p<1$, is there a good upper bound for $ \sum^{k}_{i=0} \binom{n}{i} \left(\frac{p}{1-p}\right)^{i}$ in terms of $n,k,p$ ?
There is of course the trivial bound without capturing the dependence on $k$, $  \left(1+\frac{p}{1-p}\right)^{n} = \left(\frac{1}{1-p}\right)^{n} \geq  \sum^{k}_{i=0} \binom{n}{i} \left(\frac{p}{1-p}\right)^{i}$.
 A: Let $x=p/(1-p)$. Assuming $k\leq n$, your sum can be evaluated in terms of the hypergeometric function as
$$
S:=\sum_{i=0}^k\binom{n}{i}x^i=(x+1)^n-x^{k+1} \binom{n}{k+1} F\left({1,k-n+1\atop k+2};-x\right).
$$
We can see that when $k=n$ the second term vanishes leaving your trivial upper bound; thus our goal will be to make the second term smaller without making it zero. Using transformation formula for the hypergeometric function we can write
$$
F\left({1,k-n+1\atop k+2};-x\right)=(1+x)^{-1}F\left({1,n+1\atop k+2};\frac{x}{x+1}\right)>(1+x)^{-1},
$$
since $F(1,n+1; k+2;z)$ is increasing on $z\in(0,1)$ and $F(1,n+1; k+2;0)=1$. Hence,
$$
S\leq (x+1)^n-x^{k+1}(1+x)^{-1} \binom{n}{k+1},
$$
with equality when $k=n$. Your application will determine if this upper bound is good or not. Using some trial values in Mathematica shows that this bound is almost identical to $S$ for some parameters and can be much larger for others.
For example, denoting $S^\ast$ as the upper bound above we have
$$
(p,n,k)=(0.0662009, 5, 1)\implies (S,S^\ast)=(1.35447, 1.36149),
$$
while
$$
(p,n,k)=(0.628947, 5, 2)\implies (S,S^\ast)=(38.2065, 124.103).
$$
At the very least, this upper bound does as good or better than the trivial bound.
