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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. $\endgroup$ – Pete L. Clark Mar 3 '12 at 0:41
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    $\begingroup$ Try substituting $n = e^{\log n}$... $\endgroup$ – TMM Mar 3 '12 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 Mar 3 '12 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 Nov 19 '12 at 23:38
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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 Oct 21 '14 at 14:37
  • $\begingroup$ @DigitalBrain: Glad you like it :-) $\endgroup$ – Aryabhata Oct 26 '14 at 1:42
  • $\begingroup$ spelling of aryabhata is aryabhatta! $\endgroup$ – RE60K Mar 15 '15 at 13:39
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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 Mar 17 '15 at 1:41
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    $\begingroup$ @Subhadeep: Wow! Thanks!... $\endgroup$ – Aryabhata Apr 11 '16 at 20:10
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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 Mar 3 '12 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 Mar 12 '14 at 23:43
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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 Mar 3 '12 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. $\endgroup$ – Martin Sleziak Jun 7 '12 at 9:26
  • $\begingroup$ BTW similar argument is given at PlanetMath - I found the link in comments in the other question. $\endgroup$ – Martin Sleziak Jun 7 '12 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. :-) $\endgroup$ – user 1357113 Jun 7 '12 at 9:52
  • $\begingroup$ Uniqueness and all notwithstanding, this approach is incomplete for the reason explained by @MartinSleziak. $\endgroup$ – Did Oct 6 '15 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. $\endgroup$ – user 1357113 Oct 6 '15 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. $\endgroup$ – Martin Sleziak Nov 30 '17 at 20:53
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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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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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