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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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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. – Pete L. Clark Mar 3 '12 at 0:41
Try substituting $n = e^{\log n}$... – TMM Mar 3 '12 at 0:41
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. – Maciek Mar 3 '12 at 0:50
Related (though I'm not sure about being a duplicate):… – JavaMan Nov 19 '12 at 23:38
up vote 46 down vote accepted

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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I wish I could upvote a hundred times! – user171358 Oct 21 '14 at 14:37
@DigitalBrain: Glad you like it :-) – Aryabhata Oct 26 '14 at 1:42
spelling of aryabhata is aryabhatta! – ADG Mar 15 '15 at 13:39
@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. – Aryabhata Mar 17 '15 at 1:41
@Subhadeep: Wow! Thanks!... – Aryabhata Apr 11 at 20:10

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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This proof is really trustful since you have used only elementary operations which are usually proved this result. – checkmath Mar 3 '12 at 1:53

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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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. – Steve Kass Mar 12 '14 at 23:43

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

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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I think u are using too much assumptions like continuity of $\ln t$ that need to be proved after this elementary proofs. – checkmath Mar 3 '12 at 1:32

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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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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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. – Martin Sleziak Jun 7 '12 at 9:26
BTW similar argument is given at PlanetMath - I found the link in comments in the other question. – Martin Sleziak Jun 7 '12 at 9:45
@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. :-) – user 1618033 Jun 7 '12 at 9:52
Uniqueness and all notwithstanding, this approach is incomplete for the reason explained by @MartinSleziak. – Did Oct 6 '15 at 16:53
@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. – user 1618033 Oct 6 '15 at 19:23

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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