# Proof of $\lim_{n\to \infty} \sqrt[n]{n}=1$

Thomson et al. provide a proof that $$\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$$ in this book (page 73). It has to do with using an inequality that relies on the binomial theorem: I have an alternative proof that I know (from elsewhere) as follows.

Proof.

\begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align}

Then using this, I can instead prove: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} &= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline & = \exp{0} \newline & = 1 \end{align}

On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \end{align}

I know such an identity would hold for bounded $$n$$ but I'm not sure I can use this identity when $$n\rightarrow \infty$$.

Question:

If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence $$x_n$$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n) \end{align} Or are there sequences that invalidate that identity?

(Edited to expand the last question) given any sequence $$x_n$$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n &= \exp(\log \lim_{n\rightarrow \infty} x_n) \newline &= \exp(\lim_{n\rightarrow \infty} \log x_n) \newline &= \lim_{n\rightarrow \infty} \exp( \log x_n) \end{align} Or are there sequences that invalidate any of the above identities?

(Edited to repurpose this question). Please also feel free to add different proofs of $$\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$$.

• Did you mean \begin{align} \lim_{n\rightarrow \infty} x_n = \exp(\lim_{n\rightarrow \infty}\log x_n)? \end{align} – Rasmus Mar 21 '11 at 19:04
• @Rasmus No, I didn't. But now that you mention it, I think that might be more appropriate for the question I have. – JasonMond Mar 21 '11 at 19:08
• We have $a = \exp(\log a)$ for $a \gt 0$, so what you have is basically asking whether $\lim x_n = \lim x_n$... – Aryabhata Mar 21 '11 at 19:36
• The only thing that can spoil such an identity is that $x_n$ might leave the set of definition of the function $\log$. – AD. Mar 21 '11 at 19:53
• @AD Good point. I should have thought of that. – JasonMond Mar 22 '11 at 4:08

## 8 Answers

Since $x \mapsto \log x$ is a continuous function, and since continuous functions respect limits: $$\lim_{n \to \infty} f(g(n)) = f\left( \lim_{n \to \infty} g(n) \right),$$ for continuous functions $f$, (given that $\displaystyle\lim_{n \to \infty} g(n)$ exists), your proof is entirely correct. Specifically, $$\log \left( \lim_{n \to \infty} \sqrt[n]{n} \right) = \lim_{n \to \infty} \frac{\log n}{n},$$

and hence

$$\lim_{n \to \infty} \sqrt[n]{n} = \exp \left[\log \left( \lim_{n \to \infty} \sqrt[n]{n} \right) \right] = \exp\left(\lim_{n \to \infty} \frac{\log n}{n} \right) = \exp(0) = 1.$$

• given that the limits exist – yoyo Mar 21 '11 at 21:17
• Thanks! (I never can tell whether I'm overthinking a math problem or not) – JasonMond Mar 22 '11 at 4:09

Here is one using $AM \ge GM$ to $1$ appearing $n-2$ times and $\sqrt{n}$ appearing twice.

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

i.e

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

i.e.

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

That the limit is $1$ follows.

• This "trick" is just amazing! Thanks for sharing :) – GniruT Nov 20 '15 at 12:12
• @GniruT: Thanks! Glad you liked it. – Aryabhata Nov 21 '15 at 0:28
• Mathematical tricks are the reverse of magician tricks. The magician goes "now you see it... now you don't". The mathematician goes "now you don't see it... now you do". Infinitely enjoyable. – Alecos Papadopoulos Dec 21 '20 at 22:23
• How does one say $n^{\frac{1}{n}}$ $>1$ – llecxe Apr 2 at 18:45

Here's a two-line, completely elementary proof that uses only Bernoulli's inequality:

$$(1+n^{-1/2})^n \ge 1+n^{1/2} > n^{1/2}$$ so, raising to the $2/n$ power, $$n^{1/n} < (1+n^{-1/2})^2 = 1 + 2 n^{-1/2} + 1/n < 1 + 3 n^{-1/2}.$$

I discovered this independently, and then found a very similar proof in Courant and Robbins' "What is Mathematics".

• It's worth noting that the Bernoulli inequality come from the binomial theorem. If $x>0$, then $(1+x)^n>1+nx$, and $n(n^{-1/2})=n^{1/2}$. – Aaron Aug 8 '11 at 19:45
• No. It is independent and can be easily proved by induction: True for n=1 since (1+x) >= (1+x). If true for n, then (1+x)^(n+1) = (1+x)(1+x)^n >= (1+x)(1+nx) = 1+(n+1)x+nx^2 >= 1+(n+1)x. – marty cohen Aug 8 '11 at 22:52
• Apologies. By "comes from" I meant "follows easily from" not "can only be proved using." $1+nx$ is literally the first two terms of the binomial theorem. However, having never heard of the Bernoulli inequality, I mistakenly thought that your first inequality was the Bernoulli inequality, not an application of it, which is why I didn't write a fuller explanation. You are right that it follows easily from induction, but generally speaking it is easier to remember one big theorem and its consequences than many smaller theorems. – Aaron Aug 8 '11 at 23:28
• @Aaron: But Bernoulli's inequality holds for all $x\geq-1$. There's a reason it's has its own name instead of just referring to it as first 2 terms of the binomial expansion. – user2345215 Mar 6 '14 at 22:54
• With Bernoulli, $\left(1+\frac2{\sqrt{n}}\right)^{n/2}\ge1+\sqrt{n}\implies1+\frac2{\sqrt{n}}\ge n^{1/n}$. With Binomial, $\left(1+\sqrt{\frac2n}\right)^n\ge1+n\sqrt{\frac2n}+\frac{n(n-1)}2\frac2n\implies1+\sqrt{\frac2n}\ge n^{1/n}$, which is marghially better, but at the expense of a more advanced theorem. – robjohn Apr 22 '19 at 16:22

$\sqrt[n]{n}=\sqrt[n]{1\cdot\frac{2}{1}\cdot\frac{3}{2}\dots\cdot\frac{n-1}{n-2}\cdot\frac{n}{n-1}}$ so you have a sequence of geometric means of the sequence $a_{n}=\frac{n}{n-1}$. Therefore its limit is equal to $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\frac{n}{n-1}=1$.

Let $n > 1$ so that $n^{1/n} > 1$ and we put $n^{1/n} = 1 + h$ so that $h > 0$ depends on $n$ (but we don't write the dependence explicitly like $h_{n}$ to simplify typing) Our job is done if show that $h \to 0$ as $n \to \infty$.

We have $$n = (1 + h)^{n} = 1 + nh + \frac{n(n - 1)}{2}h^{2} + \cdots$$ and hence $$\frac{n(n - 1)}{2}h^{2} < n$$ or $$0 < h^{2} < \frac{2}{n - 1}$$ It follows that $h^{2} \to 0$ as $n \to \infty$ and hence $h \to 0$ as $n \to \infty$.

Let $n$ be an integer $n>2$ and real $x>0$, the binomial theorem says $$(1+x)^n>1+nx+\frac{n(n-1)}{2}x^2$$ Let $N(x)=\max(2,1+\frac{2}{x^2})$. For $n>N(x)$, we get that $\frac{n(n-1)}{2}x^2>n$. Thus, for any $x>0$, we get that for $n>N(x)$ $$1<\sqrt[n]{n}<1+x$$ Thus, we have $$1\le\liminf_{n\to\infty}\sqrt[n]{n}\le\limsup_{n\to\infty}\sqrt[n]{n}\le 1+x$$ Since this is true for any $x>0$, we must have $$\lim_{n\to\infty}\sqrt[n]{n}=1$$

• Nitpick: It is not exactly sandwich theorem, as the limits are different ($1$ and $1+x$). You also cannot assume existence of the limit $\lim n^{1/n}$. This is easily corrected though: $\liminf n^{1/n} \ge 1$ and $\limsup n^{1/n} \le 1+x$ for arbitrary $x$ and thus $\lim n^{1/n} = 1$. +1. – Aryabhata Aug 8 '11 at 19:16
• @Aryabhata: good point. I had a more complicated proof that did properly use the Sandwich Theorem. I altered it for simplicity, but did so carelessly. I have removed the reference to the Sandwich Theorem and used $\liminf$ and $\limsup$. Thanks. – robjohn Aug 8 '11 at 19:37

The limit follows from these inequalities and the squeeze theorem: $$11$$ where the right inequality follows by keeping only the third term in the binomial expansion: $$(1+x)^n>\binom{n}{2}x^2= n,\quad \textrm{where}\quad x^2=\frac{2}{n-1}.$$

Take $$n=2^m$$

$$\lim\limits_{n \to \infty} \sqrt[n]{n} = \lim\limits_{m \to \infty} \sqrt[2^m]{2^m}= \lim\limits_{m \to \infty} 2^{\frac{m}{2^m}}=2^{\lim\limits_{m \to \infty} \frac{m}{2^m}}=2^0=1$$

This is inverted and maybe a more obvious way from the original one.