How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?

Question

I've spent the better part of this day trying to show from first principles that this sequence tends to 1. Could anyone give me an idea of how I can approach this problem?

$$ \lim_{n \to +\infty} n^{\frac{1}{n}} $$

Answer 1

You can use $\text{AM} \ge \text{GM}$.

$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge n^{1/n} \ge 1$$

$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$$

Answer 2

Let $\epsilon>0$. Choose $N$ so that ${1\over N}<\epsilon$. Noting that ${ n+1 \over n}<1+\epsilon$ for $n\ge N$: $$ N+1\le N(1+\epsilon) $$ $$ N+2 \le (N+1)(1+\epsilon)\le N (1+\epsilon)^2 $$ $$ N+3 \le (N+2)(1+\epsilon)\le N (1+\epsilon)^3 $$ $$\vdots$$ $$\tag{1} N+k \le (N+k-1)(1+\epsilon) \le N(1+\epsilon)^k. $$ Using $(1)$, we have for $n\ge N$: $$ n=N+(n-N)\le (1+\epsilon)^{n-N}N; $$ which may be written as $$ n\le B (1+\epsilon)^n, $$ where $B=N/(1+\epsilon)^N$.

Thus, for $n\ge N$ we have $$\tag {2} \root n\of { n}\le B^{1/n}(1+\epsilon). $$ Since $\lim\limits_{n\rightarrow\infty} B^{1/n}=1$, it follows from $(2)$ that $\limsup\limits_{n\rightarrow\infty} \root n\of { n}\le 1+\epsilon$.

But, as $\epsilon$ was arbitrary, we must have $\limsup\limits_{n\rightarrow\infty} \root n\of {n}\le 1 $.

Since, obviously, $\liminf\limits_{n\rightarrow\infty} \root n\of {n}\ge 1 $, we have $\lim\limits_{n\rightarrow\infty} \root n\of {n}= 1 $, as desired.

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One could also argue as follows:

Note $\root n\of n>1$ for $n>1$. For $n>1$, write $\root n\of n=1+c_n$ for some $c_n>0$. Then, by the Binonial Theorem we have, for $n>1$, $$\textstyle n=1 +nc_n+{1\over2} n(n-1)c_n^2+\cdots\ge 1+{1\over2}n(n-1)c_n^2; $$ whence $$ n-1\ge\textstyle {1\over2}n(n-1)c_n^2. $$ So, $c_n^2\le {2\over n}$ for $n>1$; whence $$ 0<\root n\of n -1=c_n\le \sqrt{2/n} $$ for $n>1$, and the result follows.

Answer 3

Fix $ \epsilon > 0 $. Then $\displaystyle \frac{(1+ \epsilon)^n}{n} \to \infty$ by the ratio test, so for all but a finite number of $n$ we have $ 1 < \displaystyle \frac{(1+ \epsilon)^n}{n},$ which can be rearranged to $\sqrt[n]{n} < 1+\epsilon .$ Thus $\sqrt[n]{n} \to 1.$

Answer 4

Applying the Binomial Theorem, we can say $$ \begin{align} \left(1+\sqrt{\frac2n}\right)^n &\ge\color{#C00}{1}+\overbrace{\binom{n}{1}\sqrt{\frac2n}}^{\sqrt{2n}}+\overbrace{\ \binom{n}{2}\frac2n\ }^{\color{#C00}{n-1}}\tag{1a}\\[6pt] &\ge\color{#C00}{n}\tag{1b} \end{align} $$ Therefore, $$ 1\le n^{1/n}\le1+\sqrt{\frac2n}\tag2 $$ to which we can apply the Squeeze Theorem.

Answer 5

$$\lim_{n \rightarrow \infty} n^{1/n} = \lim_{n \rightarrow \infty} e^{\frac{1}{n} \ln n} = e^{\lim_{n \rightarrow \infty} \frac{1}{n} \ln n}$$

With L'Hôpital's rule you can prove that $\lim_{n \rightarrow \infty} \frac{1}{n} \ln {n} = 0$. Thus, $\lim_{n \rightarrow \infty} n^{1/n} = e^0 = 1$.

Answer 6

Let's see a very elementary proof. Without loss of generality we proceed replacing $n$ by $2^n$ and get that: $$ 1\leq\lim_{n\rightarrow\infty} n^{\frac{1}{n}}=\lim_{n\rightarrow\infty} {2^n}^{\frac{1}{{2}^{n}}}=\lim_{n\rightarrow\infty} {2}^{\frac{n}{{2}^{n}}}\leq\lim_{n\rightarrow\infty} {2}^{\raise{4pt}\left.n\middle/\binom{n}{2}\right.}=2^0=1$$

By Squeeze Theorem the proof is complete.

Answer 7

Let $x_{n} = n^{\frac{1}{n}} - 1$. Then

$$ (x_{n}+1)^{n} = n.$$

By binomial expansion, you can deduce that

$$ x_{n} < \frac{2}{n-1}$$

which goes to zero and hence you have your result.

Answer 8

From binomial theorem, $(1+\epsilon)^n > (n(n-1)\epsilon^2)/2$ when $\epsilon > 0$. For any given small $\epsilon > 0$, when $n > 2/\epsilon^2 + 1$, $(1+\epsilon)^n > (n(n-1)\epsilon^2)/2 > n$, which means $n^{1/n} < 1 + \epsilon$.

Answer 9

It is totally basic and fun to do it this way:

We will prove that

$$\lim_{n \to \infty} \sqrt[n]{\frac{n}{2^n}} = \frac{1}{2}$$

from which your limit follows because

$$\lim_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = \frac{1}{2}$$

Replace $$n=2^{2^m}$$

$$\lim_{m \to \infty} \sqrt[2^{2^m}]{\frac{2^{2^m}}{2^{2^{2^m}}}} = \lim_{m \to \infty} \sqrt[2^{2^m}]{\frac{1}{2^{2^{2^m}-2^m}}} =$$

$$\lim_{m \to \infty} \frac{1}{2^{1-\frac{2^m}{2^{2^m}}}} = \lim_{m \to \infty} \frac{1}{2^{1-\frac1{2^{2^m-m}}}}$$

Now $\lim\limits_{m \to \infty} 2^m-m \to \infty$ because the difference between two successive terms $(2^{m+1}-(m+1))-(2^{m}-(m))=2^m-1$ tends to infinity meaning

$$\lim_{m \to \infty} 2^{2^m-m} = \infty $$ $$\lim_{m \to \infty} \frac1{2^{2^m-m}}=0$$

$$\lim_{m \to \infty} \frac{1}{2^{1-\frac1{2^{2^m-m}}}}=\frac{1}{2}$$

Your

$$\lim\limits_{n \to \infty} n^{\frac1{n}}=1$$ follows.

Answer 10

We know that

$$\liminf \frac{a_{n+1}}{a_n}\le \liminf (a_n)^{1/n}\le \limsup(a_n)^{1/n} \le \limsup \frac{a_{n+1}}{a_n}$$

if $(a_n)$ is a bounded sequence of positive real numbers. Take $a_n = 1/n$ and we have $\lim n^{1/n}=1 $

Answer 11

I want to add a proof that is based in the fact that $\sum \frac{a^k}{k!}=e^a$. In my opinion I found this fact easier to prove that $AM\ge GM$ inequality, thus from my point of view this is more basic. More over: we dont suppose or guess that the limit is $1$.

First suppose we proved that $\sqrt[n]{n}>\sqrt[n+1]{n+1}$ for $n\ge 2$, what is easy to do IMHO. From this proof we get very important information: the sequence $(\sqrt[n]{n})$ is decreasing and bounded below by $1$.

Then, by the monotone convergence theorem, exists some $L$ such that for all $\epsilon>0$ exists $N\in\Bbb N$ such that

$$|\sqrt[n]{n}-L|<\epsilon,\forall n\ge N$$

Suppose that $L>1$. From the square root of $2$ we know that $L<2$, in particular $L=1+a$ with $1>a>0$. Then it must be the case that $\sqrt[n]{n}-L>0$ because the sequence is bounded below by $L$. Then

$$\sqrt[n]{n}-1-a>0\iff \sqrt[n]{n}>1+a\iff n>(1+a)^n\iff 1>\frac1n (1+a)^n$$

for all $n\in\Bbb N$. Expanding the RHS we have that

$$\frac1n(1+a)^n=\frac1n\sum_{k=0}^n\binom{n}{k}a^k=\sum_{k=0}^{n}\frac{(n-1)!}{(n-k)!}\cdot\frac{a^k}{k!}\ge\sum_{k=0}^{n-1}\frac{a^k}{k!}$$

Then taking limits we have that

$$1\ge\lim_{n\to\infty}\frac1n(1+a)^n\ge\lim_{n\to\infty}\sum_{k=0}^{n-1}\frac{a^k}{k!}=e^a>1$$

what is a contradiction. Thus $L=1$.$\Box$

Answer 12

I posted a similar question and got a great answer that I don't want to be deleted with my question, so I'm posting that great answer here.

By

@Dr. Sundar

To show that $$ \lim\limits_{n \rightarrow \infty} \ a_n = 1 $$ where $$ a_n = n^{1 \over n} $$

Let $0 < \epsilon < 1 $ be given.

We note that $$ |n^{1 \over n} - 1 | < \epsilon \iff > - \epsilon < n^{1 \over n} - 1 < \epsilon \iff 1 - \epsilon < n^{1 \over n} < 1 + \epsilon $$ which is equivalent to $$ (1 - \epsilon)^n > < n < (1 + \epsilon)^n \tag{1} $$

Let $n \geq 1$ be any integer.

Clearly, $$ (1 - \epsilon)^n < 1 \leq n $$

This proves the left part of (1).

To prove the right part of (1), we note that $$ ( 1 + \epsilon )^n = > \sum\limits_{k = 0}^n \ \left( \matrix{n \cr > k \cr} \right) \ \epsilon^k > \left( \matrix{n \cr > 2 \cr} \right) \ \epsilon^2 > = {n (n - 1) \over 2} \ \epsilon^2 $$

We define $m = \lceil{ { 2 \over \epsilon^2} + 1 \rceil}$.

We choose $n > m$. Then we have $$ (1 + \epsilon)^n > {n (n - 1) \over > 2} \ \epsilon^2 > n $$

Thus, we have shown that for all $n > m$, (1) is true.

This completes the proof. $ \ \ \ \ \ \ \blacksquare$

Answer 13

You can estimate \begin{eqnarray} 1 & \leq & n^{\frac{1}{n}} = {e^{\ln(n)}}^{\frac{1}{e^{\ln(n)}}} = e^{2 \frac{1}{1!} \big(\frac{1}{2}\ln(n)\big)^1 e^{-\ln(n)}} \leq e^{2 \sum_{k=0}^\infty \frac{1}{k!}\big(\frac{1}{2}\ln(n)\big)^k e^{-\ln(n)}} \\ & = & e^{2 e^{\frac{1}{2} \ln(n)}e^{-\ln(n)}} = e^{2e^{-\frac{1}{2}\ln(n)}} \rightarrow 1 \ , \end{eqnarray} as $n \rightarrow \infty$. Increasingness of $e^x$, continuity of $e^x$ and other basic properties of $e^x$ and $\ln(x)$ are assumed. Hence the limit in the question exists and equals to $1$.