monotonicity of the sequence ${(1+x/n)}^n$

Theorem 1.

Let $x$ be a real number and let $n$ be an integer such that $n\mathrm{>}\mathrm{0}$ and $n\mathrm{+}x\mathrm{>}\mathrm{0}$. Then

Proof.

We begin by dividing the two expressions to be compared:

${\displaystyle \frac{{\left(\frac{n+x+1}{n+1}\right)}^{n+1}}{{\left(\frac{n+x}{n}\right)}^n}}$ $={\displaystyle \frac{n+x+1}{n+1}}\cdot {\left({\displaystyle \frac{n(n+x+1)}{(n+x)(n+1)}}\right)}^n$

$={\displaystyle \frac{n+x+1}{n+1}}\cdot {\left({\displaystyle \frac{n^2+nx+n}{n^2+nx+n+x}}\right)}^n$

$={\displaystyle \frac{n+x+1}{n+1}}\cdot {\left(1-{\displaystyle \frac{x}{n^2+nx+n+x}}\right)}^n$

Now, when $x>0$, we have

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

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

${\left(1-{\displaystyle \frac{x}{n^2+nx+n+x}}\right)}^n$ $>1-{\displaystyle \frac{nx}{n^2+nx+n+x}}$

$={\displaystyle \frac{n^2+n+x}{n^2+nx+n+x}}$

Hence, we have

${\displaystyle \frac{{\left(\frac{n+x+1}{n+1}\right)}^{n+1}}{{\left(\frac{n+x}{n}\right)}^n}}$ $>{\displaystyle \frac{(n+x+1)(n^2+n+x)}{(n+1)(n^2+nx+n+x)}}$

$={\displaystyle \frac{n^3+2n^2+n+n^{2}x+2nx+x+x^2}{n^3+2n^2+n+n^{2}x+2nx+x}}$

Note that the numerator is greater than the denominator because it contains every term contained in the denominator and an extra term $x^2$. Hence this ratio is larger than $1$; multiplying out, we obtain the inequality 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