proof of limit of nth root of n

In this entry, we present a self-contained, elementary proof of the fact that  ${lim}_{n\to \mathrm{\infty }}{n}^{1/n}=1$. We begin by with inductive proofs of two integer inequalities — real numbers will not enter until the very end.

Lemma 1.

For all integers $n$ greater than or equal to $\mathrm{5}$,

$$2^n>n^2$$

Proof.

We begin with a few easy observations. First, a bit of arithmetic:

$$2^5=32>25=5^2$$

Second, some algebraic manipulation of the inequality $n>4$:

$n-1$ $>3$

${(n-1)}^2$ $>9$

${(n-1)}^2$ $>2$

$n^2-2n+1$ $>2$

$2n^2$ $>n^2+2n+1$

$2n^2$ $>{(n+1)}^2$

These observations provide us with the makings of an inductive proof. Suppose that $2^n>n^2$ for some integer $n\ge 5$. Using the inequality we just showed,

$$2^{n+1}=2\cdot 2^n>2n^2>{(n+1)}^2.$$

Snce $2^5>5^2$ and $2^n>n^2$ implies that $2^{n+1}>{(n+1)}^2$ when $n\ge 5$ we conclude that $2^n>n^2$ dor all $n\ge 5$. ∎

Lemma 2.

For all integers $n$ greater than or equal to $\mathrm{3}$,

$$n^{n+1}>{(n+1)}^n$$

Proof.

We begin by noting that

$$3^4=81>64=4^3.$$

Next, we make assume that

$${(n-1)}^n>n^{(n-1)}.$$

for some $n$. Multiplying both sides by $n$:

$$n{(n-1)}^n>n^n.$$

Multiplying both sides by ${(n+1)}^n$ and making use of the identity $(n+1)(n-1)=n^2-1$,

$$n{(n^2-1)}^n>n^n{(n+1)}^n.$$

Since $n^2>n^2-1$, the left-hand side is less than $n^{2n+1}$, hence

$$n^{2n+1}>n^n{(n+1)}^n.$$

Canceling $n^n$ from both sides,

$$n^{(n+1)}>{(n+1)}^n.$$

Hence, by induction, $n^{(n+1)}>{(n+1)}^n$ for all $n\ge 3$. ∎

Theorem 1.

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{1/n}=1$$

Proof.

Consider the subsequence where $n$ is a power of $2$. We then have

$${(2^m)}^{(1/2^m)}=2^{m/2^m}.$$

By lemma 1, when $m\ge 5$. Hence, . Since ${lim}_{m\to 0}{2}^{1/m}=1$, and ${(2^m)}^{1/2^m)}>1$, we conclude by the squeeze rule that

$$\underset{m\to 0}{lim}{(2^m)}^{1/2^m}=1.$$

By lemma 2, the sequence $\{n^{1/n}\}$ is decreasing. It is clearly bounded from below by $1$. Above, we exhibited a subsequence which tends towards $1$. Thus it follows that

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{1/n}=1.$$

∎

Title	proof of limit of nth root of n
Canonical name	ProofOfLimitOfNthRootOfN
Date of creation	2014-02-28 7:21:31
Last modified on	2014-02-28 7:21:31
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	19
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 30-00
Classification	msc 12D99