I am having some trouble showing the following estimation:
Let $\mu \in (0, 1)$ be an absolute constant. There exists an absolute constant $\eta \in (0, 1)$ small enough such that the number of subsets of $\{ 1, \cdots, N \}$ with cardinality at least $(1 - 4\eta^2) N$ is at most $\mu^{-N/4}$ for all $N \in \mathbb{N}$.
My attempt is the following:
We know this is equivalent to showing the estimation for sum of binomial coefficients. That is, we want to find $\eta \in (0, 1)$ small enough such that$$ \sum_{j = (1 - 4\eta^2)N} ^N \binom{N}{j} \leq \mu^{-N/4}.$$In particular, this is equivalent to saying finding a portion size (absolute constant) $\alpha \in (0, 1)$ small enough such that$$\sum_{j = \alpha N} ^N \binom{N}{j} \leq \mu^{-N/4}.$$We may roughly write the sum as$$\sum_{j = 0} ^N \binom{N}{j} - \sum_{j = 0} ^{\alpha N} \binom{N}{j}.$$According to the classical inequality on sum of binomials, we may upper bound the first term above by $\left( \frac{eN}{N} \right)^N = e^N$ and lower the second term above by $\binom{N}{\alpha N}$. In particular, we can write the estimation$$\sum_{j = 0} ^N \binom{N}{j} - \sum_{j = 0} ^{\alpha N} \binom{N}{j} \leq e^N - \binom{N}{\alpha N}.$$Thus now we need to show$$e^N - \binom{N}{\alpha N} \leq \mu^{-N/4}$$for all $N$. I am stuck at this step. Is there an easy way to estimate what the absolute constant $\alpha$ is supposed to be here?