Binomial approximation with additional term

Question

From Stirling's approximation we know that $$\binom{2n}{n}\sim \frac{2^{2n}}{\sqrt{\pi n}}.$$

Is it true that for some constant $c$ and for $n$ large enough, $$\binom{2n}{n+10}\geq c\cdot\frac{2^{2n}}{\sqrt{\pi n}}?$$

Answer 1

For any $n>10$,

$$\frac{\binom{2n}{n+10}}{\binom{2n}{n}} = \frac{n!^2}{(n+10)!(n-10)!} $$ and the limit of the RHS as $n\to +\infty$ is just $1$. It follows that for any $c\in(0,1)$ the inequality $$ \binom{2n}{n+10}\geq c\cdot\frac{4^n}{\sqrt{\pi n}} $$ holds for any $n$ large enough.