How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$? 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?
$$
\lim_{n \to +\infty} n^{\frac{1}{n}}
$$
 A: You can use $\text{AM} \ge \text{GM}$.
$$\frac{1 + 1 + \dots + 1 + \sqrt{n} + \sqrt{n}}{n} \ge n^{1/n} \ge 1$$
$$ 1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge n^{1/n} \ge 1$$
A: $$\lim_{n \rightarrow \infty} n^{1/n} = \lim_{n \rightarrow \infty} e^{\frac{1}{n} \ln n} = e^{\lim_{n \rightarrow \infty} \frac{1}{n} \ln n}$$
With L'Hôpital's rule you can prove that $\lim_{n \rightarrow \infty} \frac{1}{n} \ln {n} = 0$. Thus, $\lim_{n \rightarrow \infty} n^{1/n} = e^0 = 1$.
A: 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.
A: 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.
A: 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.
A: 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.  


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.
A: 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 $\sqrt[n]{n} \to 1.$  
A: 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 $
A: 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 $n^{1/n} < 1 + \epsilon$.
A: 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 $1$.
A: 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 $L=1$.$\Box$
A: 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.
A: 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$

