proof of limit of nth root of n

In this entry, we present a self-contained, elementaryproof of the fact that  ${lim}_{n\to \mathrm{\infty }}{n}^{1/n}=1$.We begin by with inductive proofs of two integerinequalities — real numbers will not enter untilthe 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 aninductive proof. Suppose that $2^n>n^2$ for someinteger $n\ge 5$. 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 $2^n>n^2$ dorall $n\ge 5$.∎

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 makinguse 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 isless 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 concludeby 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. Itis clearly bounded from below by $1$. Above, we exhibiteda subsequence which tends towards $1$. Thus it follows that

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

∎