Recall Stirling's formula, then determine a real number $c > 0$ such that $$\binom { 2 n } { n } \underset { n \rightarrow + \infty } { \sim } c \frac { 4 ^ { n } } { \sqrt { n } }$$
Recall Stirling's formula, then determine a real number $c > 0$ such that
$$\binom { 2 n } { n } \underset { n \rightarrow + \infty } { \sim } c \frac { 4 ^ { n } } { \sqrt { n } }$$