# proof of limit of nth root of n

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

###### Lemma 1.

For all integers $n$ greater than or equal to $5$,

 $2^{n}>n^{2}$
###### 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 $n>4$:

 $\displaystyle n-1$ $\displaystyle>3$ $\displaystyle(n-1)^{2}$ $\displaystyle>9$ $\displaystyle(n-1)^{2}$ $\displaystyle>2$ $\displaystyle n^{2}-2n+1$ $\displaystyle>2$ $\displaystyle 2n^{2}$ $\displaystyle>n^{2}+2n+1$ $\displaystyle 2n^{2}$ $\displaystyle>(n+1)^{2}$

These observations provide us with the makings of an inductive proof. Suppose that $2^{n}>n^{2}$ for some integer $n\geq 5$. 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 $n\geq 5$ we conclude that $2^{n}>n^{2}$ dor all $n\geq 5$. ∎

###### Lemma 2.

For all integers $n$ greater than or equal to $3$,

 $n^{n+1}>(n+1)^{n}$
###### 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 $n\geq 3$. ∎

###### Theorem 1.
 $\lim_{n\to\infty}n^{1/n}=1$
###### 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, $m/2^{m}<1/m$ when $m\geq 5$. Hence, $(2^{m})^{1/2^{m}}<2^{1/m}$. Since $\lim_{m\to 0}2^{1/m}=1$, and $(2^{m})^{1/2^{m})}>1$, we conclude by the squeeze rule that

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

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

 $\lim_{n\to\infty}n^{1/n}=1.$

Title proof of limit of nth root of n ProofOfLimitOfNthRootOfN 2014-02-28 7:21:31 2014-02-28 7:21:31 rspuzio (6075) rspuzio (6075) 19 rspuzio (6075) Proof msc 30-00 msc 12D99