Problem proving inequality (4^n)/(n+1)<=((2n)!)/((n!)^2) I skip the base case n=0 because it's obvious. I know that this is very equivalent to: Prove by induction: (2^(2n))/(n+1)<((2n)!)/((n!)^2),n>1

rochioblogmz 2022-10-23 Answered
Problem proving inequality 4 n n + 1 ( 2 n ) ! ( n ! ) 2
I skip the base case n = 0 because it's obvious.
I know that this is very equivalent to: Prove by induction: 2 2 n n + 1 < ( 2 n ) ! ( n ! ) 2 , n > 1
But I try to learn some tricks so maybe you can help me?
Assumption: 4 n n + 1 ( 2 n ) ! ( n ! ) 2 4 n ( n + 1 ) ( 2 n ) ! ( n ! ) 2 for some n.
Step: n n + 1
4 n + 1 = ( n + 2 ) ( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2
I begin with manipulating the LHS:
4 n + 1 = 4 4 n ? ? ?
? ? ? ( 2 n ) ! ( 2 n + 2 ) ( 2 n + 1 ) ( n + 1 ) ( n ! ( n + 1 ) )
= ( n + 1 ) ( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2
Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.
Can you help me?
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Answers (1)

scranna0o
Answered 2022-10-24 Author has 16 answers
4 n + 1 ( n + 2 ) < ( 2 n ) ! ( n ! ) 2 4 ( n + 1 ) ( n + 2 )
Now
( 2 n ) ! ( n ! ) 2 4 ( n + 1 ) ( n + 2 ) = ( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2 4 ( n + 1 ) 3 ( n + 2 ) ( 2 n + 1 ) ( 2 n + 2 )
So
4 ( n + 1 ) 3 ( n + 2 ) ( 2 n + 1 ) ( 2 n + 2 ) < 1
implies
( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2 4 ( n + 1 ) 3 ( n + 2 ) ( 2 n + 1 ) ( 2 n + 2 ) < ( 2 n + 2 ) ! ( ( n + 1 ) ! ) 2
, for n > 1
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