Given a probability $p \in (0,1)$ and parameter $\alpha \in (0,1)$, is there an entropy-based proof which yields a good upper bound for the sum
$$\sum\_{\ell = 0}^{\alpha n} \binom{n}{\ell}p^\ell(1-p)^{n-\ell}$$
when $n$ is large?


When $p = 1/2$, there is very simple proof (for example, see section 3.1 of [this paper](https://arxiv.org/pdf/1406.7872.pdf)) which upper bounds the above quantity by
$$2^{(H(\alpha) - 1)n}$$
when $H(\cdot)$ denotes the [binary entropy function](https://en.wikipedia.org/wiki/Binary_entropy_function).


Is there a proof using similar techniques which gives a bound for the more general sum above (which can be interpreted as the CDF of a binomial distribution with parameter $p$)?


I'd also be interested in other proofs for bounds on the above sum. The appropriate bound has already noted in [this answer](https://mathoverflow.net/a/93756/138628), but doesn't sketch out a proof establishing this result.



Is there an entropy proof for bounding a weighted sum of binomial coefficients?
Yes, if $\alpha<p$ (if $\alpha>p$, the sum is almost 1). To see this, write
$$
\sum\_{\ell = 0}^{\alpha n} \binom{n}{\ell}p^\ell(1-p)^{n-\ell}\leqslant t^{-\alpha n}(pt+(1-p))^n
$$
for every $t\in (0,1]$. Choose a positive $t=t\_0$ for which RHS is minimal possible, taking the logarithmic derivative equal to 0 we get $-\alpha/t\_0+p/(pt\_0+1-p)=0$, $-\alpha p t\_0-\alpha(1-p)+pt\_0=0$, $pt\_0(1-\alpha)=\alpha(1-p)$, $t\_0=\frac{\alpha(1-p)}{p(1-\alpha)}$. We see that if $\alpha\leqslant p$, this $t\_0$ is indeed in $(0,1]$, thus we get the upper bound $\theta^n$, where
$$
\theta=\frac{p^\alpha(1-p)^{1-\alpha}}{\alpha^\alpha (1-\alpha)^{1-\alpha}}.
$$


