PlanetMath: monotonicity of the sequence $(1 + x/n)^n$

Proof. We begin by dividing the two expressions to be compared:

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

Now, when

, we have

$\displaystyle 0 < {x \over n^2 + n x + n + x} < 1$

whilst, when

and

, we have,

$\displaystyle {x \over n^2 + n x + n + x} < 0 .$

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

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

Hence, we have

$\displaystyle {\left( {n + x + 1 \over n + 1} \right)^{n+1} \over \left( {n + x \over n} \right)^n}$	$\displaystyle > {(n + x + 1) (n^2 + n + x ) \over (n + 1) (n^2 + n x + n + x)}$
	$\displaystyle = {n^3 + 2 n^2 + n + n^2 x + 2 n x + x + x^2 \over n^3 + 2 n^2 + n + n^2 x + 2 n x + x}$

Note that the numerator is greater than the denominator because it contains every term contained in the denominator and an extra term

. Hence this ratio is larger than

; multiplying out, we obtain the inequality which was to be demonstrated. $\qedsymbol$