How to show that $\sum\limits_{k=0}^{\lfloor0.999n\rfloor}\binom{2n}{k} < \binom{2n}{n} $ holds for large n It seems logical to me since $\binom{2n}{n}$ is in the middle of the row in pascal triangle; therefore, the largest, and for large n the sum adds only the small ones on the left. But I do not have any idea how to show it. I tried some basic approximations but nothing worked. Thanks for your help. 
 A: We use Chernoff bound (alternatively Hoeffding's inequality) for binomial distribution.
With $X \sim B(2n, 1/2)$, we have by Chernoff bound, 
$$
P(X\leq 0.999n) \leq \exp\left( -10^{-6} n \right).
$$
On the other hand, by Stirling's formula,
$$
P(X=n) \sim \frac 1{2^{2n}} \frac{\sqrt{2\pi \cdot 2n} \left( \frac {2n}e \right)^{2n}}{(\sqrt{2\pi n} )^2 \left(\frac ne\right)^{2n} } \sim \frac 1{\sqrt{\pi n}}.
$$
Thus, for sufficiently large $n$, we have 
$$
P(X\leq 0.999n)< P(X=n).$$
Elementary way
We can avoid Chernoff bound and Stirling's formula, and prove this in a completely elementary way. As @Nate suggested, we apply Chebyshev inequality. 
$$
P(X\leq 0.999n)=\frac 12 P( |X- n|\geq 0.001n ) \leq \frac 12 \cdot \frac {\mathrm{Var}(X)}{(0.001n)^2} = \frac {250000 }n.
$$
For the central probability, use
$$
P(X=n)=\frac1{2^{2n}} \frac{(2n)!}{n!n!}=\frac{ 1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}.
$$
Then by squaring, we have
$$
P(X=n)^2 = \left( \frac 12\right)^2 \left( \frac 34 \right)^2 \cdots \left(\frac{2n-1}{2n}\right)^2.$$
Then replace $\left(\frac 34\right)^2 $ by $\frac 23 \cdot \frac 34$, and $\left(\frac 56\right)^2 $ by $\frac 45\cdot \frac 56$, $\ldots$, lastly $\left(\frac{2n-1}{2n}\right)^2$ by $\frac{2n-2}{2n-1}\cdot\frac{2n-1}{2n}$. This gives
$$
P(X=n)^2 \geq \frac 12 \cdot \frac 1{2n}=\frac1{4n}.
$$
Now, we have
$$
P(X=n)\geq \frac 1{2\sqrt n}.
$$
Therefore, for $n> 500000^2$, we have
$$
P(X\leq 0.999n)< P(X=n).
$$
