I was recently learning about binomial coefficients and was wondering about how to disprove 2nCn (or the central binomial coefficient) not being lower-bounded by 4^n; in other words:
Some extremely generous bounds can be easily constructed, such as the following:
I sought to prove by contradiction, so to assume:
Clearly, c 1 cannot exist, since 1/(2n + 1) approaches 0 as n approaches infinity. It can also be seen that c 2 must reside in (0, 1]. And... I'm stuck. Intuitively, it seems rather obvious that c 2 cannot exist.
I am aware a similar question has been asked here , but there wasn't really a proof provided. I'm also aware that you could prove the limit of 2nCn/4 n approaches 0 as n approaches infinity, but I was wondering if there was another way to do so - particularly, by proving that c 2 cannot exist.
For all n, the constant c 2 would have to be upper bounded by
2n choose n (2n)!
----------- = -------------
4^n 2^n n! 2^n n!
(2n)!
= -------------
(2n)!! (2n)!!
(2n-1)!!
= --------
(2n)!!
n (2i-1)
= product ------
i=1 2i
n
= product (1 - 1/(2i))
i=1
n
≤ product exp(-1/(2i)) [since 1 + x ≤ exp(x)]
i=1
n
= exp(sum -1/(2i))
i=1
≤ exp(-ln(n+1)/2) [since sum ≤ integral of increasing fn]
= 1/√(n+1),
hence it cannot be positive.
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