# 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} Mar 21, 2011 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. Mar 21, 2011 at 19:08
• We have $a = \exp(\log a)$ for $a \gt 0$, so what you have is basically asking whether $\lim x_n = \lim x_n$... Mar 21, 2011 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 $\log$. Mar 21, 2011 at 19:53
• @AD Good point. I should have thought of that. Mar 22, 2011 at 4:08

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 :) Nov 20, 2015 at 12:12
• @GniruT: Thanks! Glad you liked it. Nov 21, 2015 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. Dec 21, 2020 at 22:23
• How does one say $n^{\frac{1}{n}}$ $>1$ Apr 2, 2021 at 18:45

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, 2011 at 21:17
• Thanks! (I never can tell whether I'm overthinking a math problem or not) Mar 22, 2011 at 4:09

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 $n(n^{-1/2})=n^{1/2}$. Aug 8, 2011 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. Aug 8, 2011 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. Aug 8, 2011 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. Mar 6, 2014 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, 2019 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. Aug 8, 2011 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, 2011 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.