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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?

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

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  • $\begingroup$ I'm not quite sure which principles are "first", but the standard method here is to take the logarithm of the limit, use L'Hopital's Rule, and then exponentiate back. $\endgroup$ Commented Mar 3, 2012 at 0:41
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    $\begingroup$ Try substituting $n = e^{\log n}$... $\endgroup$
    – TMM
    Commented Mar 3, 2012 at 0:41
  • $\begingroup$ Ah, sorry for the confusion. Basically I'm "not allowed" to use L'Hopitals rule yet. Aside from the formal definition of a limit pretty much all I can use is direct comparison and ratio tests. $\endgroup$
    – Maciek
    Commented Mar 3, 2012 at 0:50
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    $\begingroup$ Related (though I'm not sure about being a duplicate): math.stackexchange.com/questions/28348/… $\endgroup$
    – JavaMan
    Commented Nov 19, 2012 at 23:38
  • $\begingroup$ How does one say $n^{\frac{1}{n}}$ $>1$ $\endgroup$
    – llecxe
    Commented Apr 2, 2021 at 19:06

13 Answers 13

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

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

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

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    $\begingroup$ I wish I could upvote a hundred times! $\endgroup$
    – user171358
    Commented Oct 21, 2014 at 14:37
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    $\begingroup$ @DigitalBrain: Glad you like it :-) $\endgroup$
    – Aryabhata
    Commented Oct 26, 2014 at 1:42
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    $\begingroup$ @ADG: It is my name, I will spell it as I want! :-). Just kidding :-). Apparently, it is actually Aryabhata and not Aryabhatta. In fact I had it as Aryabhatta till ShreevatsaR corrected me. $\endgroup$
    – Aryabhata
    Commented Mar 17, 2015 at 1:41
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    $\begingroup$ @Subhadeep: Wow! Thanks!... $\endgroup$
    – Aryabhata
    Commented Apr 11, 2016 at 20:10
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    $\begingroup$ @llecxe: if a > 0 and b > 1 then b^a > 1. $\endgroup$
    – Aryabhata
    Commented Apr 16, 2021 at 22:25
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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.

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  • $\begingroup$ This proof is really trustful since you have used only elementary operations which are usually proved this result. $\endgroup$
    – checkmath
    Commented Mar 3, 2012 at 1:53
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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 $\sqrt[n]{n} \to 1.$

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  • $\begingroup$ For your final “thus,” you need an additional hypothesis such as the fact that for all $n>1$, $\sqrt[n]{n}>1$. The fact that $a_n<L$ for all but a finite number of $n$ doesn’t imply that $\lim_{n\rightarrow\infty}a_n=L$, as can be seen by taking $a_n=0$, for example. $\endgroup$
    – Steve Kass
    Commented Mar 12, 2014 at 23:43
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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.

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    $\begingroup$ +1 very nice RobJohn $\endgroup$ Commented Nov 28, 2020 at 23:18
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$$\lim_{n \rightarrow \infty} n^{1/n} = \lim_{n \rightarrow \infty} e^{\frac{1}{n} \ln n} = e^{\lim_{n \rightarrow \infty} \frac{1}{n} \ln n}$$

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

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  • $\begingroup$ I think u are using too much assumptions like continuity of $\ln t$ that need to be proved after this elementary proofs. $\endgroup$
    – checkmath
    Commented Mar 3, 2012 at 1:32
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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.

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    $\begingroup$ If we want to replace sequence by subsequence in the above argument, maybe we should prove first that the sequence is monotone. (Or use some other argument that the limit exists.) The fact that this sequence is decreasing for $n\ge 3$ is mentioned as a hint here. $\endgroup$ Commented Jun 7, 2012 at 9:26
  • $\begingroup$ BTW similar argument is given at PlanetMath - I found the link in comments in the other question. $\endgroup$ Commented Jun 7, 2012 at 9:45
  • $\begingroup$ @Martin Sleziak: nice. I didn't know that a similar proof is to be found on PlanetMath. I usually try to post some unique solutions, my own solutions, but they are just apparently unique because it's possible to find them in elsewhere. :-) $\endgroup$ Commented Jun 7, 2012 at 9:52
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    $\begingroup$ Uniqueness and all notwithstanding, this approach is incomplete for the reason explained by @MartinSleziak. $\endgroup$
    – Did
    Commented Oct 6, 2015 at 16:53
  • $\begingroup$ @Did Right. Alternatively, having in mind a similar idea, we see that $$1\leq\ n^{\frac{1}{n}}\le (1+\epsilon)^{\sqrt{n}/n}, \ \epsilon>0$$ for some $n\ge n_0$. Taking the limit as $n \to \infty$ we're done. $\endgroup$ Commented Oct 6, 2015 at 19:23
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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.

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    $\begingroup$ Did you mean to write $x_n^2<\frac2{n-1}$ rather than $x_n<\frac2{n-1}$? I'll add link to this answer which uses similar approach, but includes a bit more details. $\endgroup$ Commented Nov 30, 2017 at 20:53
  • $\begingroup$ Since $e^x-1\ge x$, we have $x_n=n^{1/n}-1\ge\frac{\log(n)}{n}$. Therefore, it appears that $x_n\lt\frac2{n-1}$ cannot be true for $n\ge10$. $\endgroup$
    – robjohn
    Commented Nov 29, 2020 at 9:41
  • $\begingroup$ How does one say $n^{\frac{1}{n}}$ $>1$ $\endgroup$
    – llecxe
    Commented Apr 2, 2021 at 18:36
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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 $

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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 $n^{1/n} < 1 + \epsilon$.

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    $\begingroup$ In my view, you should provide some explanation. $\endgroup$
    – 311411
    Commented Aug 26, 2021 at 19:33
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    $\begingroup$ Please add further details to expand on your answer, such as working code or documentation citations. $\endgroup$
    – Community Bot
    Commented Aug 26, 2021 at 19:33
  • $\begingroup$ There is nothing new here that has not already been said. This is basically a variation of the answers by robjohn or DavidMitra above. In general, you should only provide answers that provide a new perspective or a new approach, not a rehash of existing answers. $\endgroup$
    – PatrickR
    Commented Aug 26, 2021 at 21:25
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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.

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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 $1$.

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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 $L=1$.$\Box$

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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$

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