# 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

 $\left({n+x\over n}\right)^{n}<\left({n+1+x\over n+1}\right)^{n+1}.$
###### Proof.

We begin by dividing the two expressions to be compared:

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

Now, when $x>0$, we have

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

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

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

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

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

Hence, we have

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

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 $(1+x/n)^{n}$ MonotonicityOfTheSequence1Xnn 2013-03-22 17:01:47 2013-03-22 17:01:47 rspuzio (6075) rspuzio (6075) 18 rspuzio (6075) Theorem msc 32A05