monotonicity of the sequence (1+x/n)n

Theorem 1.

Let x be a real number and let n be an integer such that n>0 and n+x>0. Then


We begin by dividing the two expressions to be compared:

(n+x+1n+1)n+1(n+xn)n =n+x+1n+1(n(n+x+1)(n+x)(n+1))n

Now, when x>0, we have


whilst, when x<0 and n+x>0, we have,


Therefore, we may apply an inequality for differences of powers to conclude

(1-xn2+nx+n+x)n >1-nxn2+nx+n+x

Hence, we have

(n+x+1n+1)n+1(n+xn)n >(n+x+1)(n2+n+x)(n+1)(n2+nx+n+x)

Note that the numerator is greater than the denominator because it contains every term contained in the denominator and an extra term x2. Hence this ratio is larger than 1; multiplying out, we obtain the inequalityMathworldPlanetmath which was to be demonstrated. ∎

Title monotonicity of the sequence (1+x/n)n
Canonical name MonotonicityOfTheSequence1Xnn
Date of creation 2013-03-22 17:01:47
Last modified on 2013-03-22 17:01:47
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 18
Author rspuzio (6075)
Entry type Theorem
Classification msc 32A05