TITLE: Some trouble with the induction
QUESTION [7 upvotes]: Prove, that for any positive integer $n \geqslant 2$ we have the
  inequality $$ \frac{ 4^n }{ n+1 } < \frac{ (2n)! }{ (n!)^2 }.$$


For $n=2$ the inequality is true. Directly just take and prove inequality for $k+1$ problematically. So, I think we need to find the recurrence relation to one of the members of inequality. Need some hint!

REPLY [5 votes]: A simple proof is based on the observation that $\dfrac{(2n)!}{(n!)^2}$
 is the central binomial coefficient $\displaystyle{ {2n} \choose n}$.
Look at row $2n$ in the Pascal triangle. The sum of all $n+1$ terms is $2^{2n}= 4^n$. Now, the central binomial coefficient is the largest number in that row and so $4^n \le (n+1){{2n} \choose n}$.
[I've used $a_1 \le M, \dots, a_k \le M$ implies $a_1+\cdots +a_{k} \le kM$.]