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