proof of upper and lower bounds to binomial coefficient


Let 2≤k≤n be natural numbersMathworldPlanetmath. We’ll first prove the inequalityMathworldPlanetmath

(nk)≤(n⁢ek)k.

We rewrite (nk) as

(nk) = n⁢(n-1)⁢⋯⁢(n-k+1)k!
= (1-1n)⁢⋯⁢(1-k-1n)⋅nkk!

Since each of the parenthesized factors lies between 0 and 1, we have

(nk)<nkk!

Since all the terms of the series ek=∑n=0∞kn/n! are positive when k is a positive real number, each term must be smaller than the whole sum; in particular, this implies that, for any non-negative integer k, we have ek>kk/k!. Rearranging this slightly,

1<k!⁢ekkk

Multiplying this inequality by the previous inequality for the binomial coefficientMathworldPlanetmath yields

(nk)<nkk!⋅k!⁢ekkk=(n⁢ek)k

To conclude the proof we show that

∏i=1n-1(1+1i)i=nnn!⁢∀n≥2∈ℕ. (1)
∏i=1n-1(1+1i)i = ∏i=1n-1(i+1)iii
= ∏i=2nii-1(∏i=1n-1ii-1)⋅(n-1)!

Since each left-hand factor in (1) is <e, we have nnn!<en-1. Since n-i<n⁢∀ 1≤i≤k-1, we immediately get

(nk) = ∏i=2k-1(1-1i)k!<nkk!.

And from

k≤n ⇔ (n-i)⋅k≥(k-i)⋅n⁢∀ 1≤i≤k-1

we obtain

(nk) = nk⋅∏i=1k-1n-ik-i
≥(nk)k.
Title proof of upper and lower bounds to binomial coefficient
Canonical name ProofOfUpperAndLowerBoundsToBinomialCoefficient
Date of creation 2013-03-22 13:49:06
Last modified on 2013-03-22 13:49:06
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 12
Author rspuzio (6075)
Entry type Proof
Classification msc 05A10