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?

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

• 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. 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. Mar 3 '12 at 0:50
• Related (though I'm not sure about being a duplicate): math.stackexchange.com/questions/28348/… Nov 19 '12 at 23:38
• How does one say $n^{\frac{1}{n}}$ $>1$ Apr 2 '21 at 19:06

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

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

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

• I wish I could upvote a hundred times!
– user171358
Oct 21 '14 at 14:37
• @DigitalBrain: Glad you like it :-) Oct 26 '14 at 1:42
• @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. Mar 17 '15 at 1:41
• @Subhadeep: Wow! Thanks!... Apr 11 '16 at 20:10
• @llecxe: if a > 0 and b > 1 then b^a > 1. Apr 16 '21 at 22:25

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.

• This proof is really trustful since you have used only elementary operations which are usually proved this result. 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 $\sqrt[n]{n} \to 1.$

• 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. 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, $\lim_{n \rightarrow \infty} n^{1/n} = e^0 = 1$.

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

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.

• 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. Jun 7 '12 at 9:26
• BTW similar argument is given at PlanetMath - I found the link in comments in the other question. 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. :-) 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. Oct 6 '15 at 19:23

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.

• +1 very nice RobJohn Nov 28 '20 at 23:18

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.

• 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. Nov 30 '17 at 20:53
• 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 $n\ge10$.
– robjohn
Nov 29 '20 at 9:41
• How does one say $n^{\frac{1}{n}}$ $>1$ Apr 2 '21 at 18:36

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 $\lim n^{1/n}=1$

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

• In my view, you should provide some explanation. Aug 26 '21 at 19:33
– Community Bot
Aug 26 '21 at 19:33
• 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. Aug 26 '21 at 21:25

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 $L=1$.$\Box$

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.