Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that $\lim_{n \to \infty} a^{\frac{1}{n}} = 1$ if $a >0$. In my textbook, we are given a suggestion to let $a^{\frac{1}{n}} = (1+h_n)$ and then show that the $h_n$ term goes to zero using a Theorem that states the following conditions:

$\lim_{n \to \infty} = \infty,$ if $a >1, 1,$ if $a = 1,$ and 0, if $\vert a \vert < 1$. I apologize for not formatting the above cases appropriately; I could not figure out how to use a giant left brace to group them altogether.

If I rewrite it as $a^{n^{-1}}$, then I thought that this might help, but I don't think the Binomial Theorem would then apply since the exponent is negative. Furthermore, the problem is only concerned with the positive, real $n$-th roots. I am quite stuck on this problem and any suggestions or advice would be greatly appreciated.

I am using the textbook Introduction to Analysis by Arthur Mattuck.

share|improve this question
The binomial theorem does apply to negative exponents. For example, $(1+X)^{-1} = 1 + (-1)X + (-1)(-2)/2 X^2 + (-1)(-2)(-3)/3! X^3 + ... = 1 - X + X^2 - X^3 + ...$ –  Cocopuffs Sep 9 '13 at 3:39
Oh, yes you are absolutely correct. I made a mistake there. Thank you for pointing that one out. –  Jamil_V Sep 9 '13 at 3:47
The error has been corrected in the post. –  Jamil_V Sep 9 '13 at 4:02
add comment

3 Answers 3

up vote 1 down vote accepted

Note: I first saw this in "What is Mathematics?" by Courant and Robbins.

You need two cases: $0 < a < 1$ and $a > 1$. These both use the motto "Always expand around zero."

These cases both use Bernoulli's inequality (if $x \ge 0$ and $n \ge 1$ then $(1+x)^n \ge 1+xn$). These also implicitly use the Archimedean axiom for the real numbers (for any positive reals $x$ and $y$ there is an integer $m$ such that $mx > y$).

If $a > 1$, let $a^{1/n} = 1+b$ where $b > 0$. Then $a =(1+b)^n \ge 1+bn $ so $b \le \frac{a-1}{n} $. We can make $b < \epsilon$ for any $\epsilon$ by choosing $n > \frac{a-1}{\epsilon} $.

If $0 < a < 1$, let $a^{1/n} = \frac1{1+b}$ where $b > 0$. Then $a = \frac1{(1+b)^n}$. As before, $(1+b)^n \ge 1+bn $, so

$\begin{align} &a \le \frac1{1+bn}\\ &\text{or}\\ &\frac1{a} \ge 1+bn\\ &\text{or}\\ &b \le \frac{1/a-1}{n}\\ &\text{or}\\ &1+b \le 1+\frac{1/a-1}{n} = \frac{n+1/a-1}{n}\\ &\text{or}\\ &\frac1{1+b} \ge \frac{n}{n+1/a-1} = 1-\frac{1/a-1}{n+1/a-1}\\ \end{align} $

We can make $\frac1{1+b} > 1-\epsilon $ by making $n+1/a-1 >\frac{1/a-1}{\epsilon} $ or $n >(\frac1{a}-1)(\frac1{\epsilon}-1) $ for any $\epsilon > 0$.

share|improve this answer
Ok, it took me a while to understand this proof. I am only a little confused on the part where you add 1 to both sides of the inequality $b \leq \frac{1/a-1}{n}$. Is this used to produce a very small $\epsilon$ when the reciprocal law is applied? –  Jamil_V Sep 11 '13 at 3:14
That's because we need to get $\frac1{1+b}$, and to get that we need $1+b$. –  marty cohen Sep 11 '13 at 4:33
Ok, I see. Then what happens in the final step when $\frac{1}{1+b} \geq \frac{n}{n+1/a-1}$? In other words, how does that become $\frac{n}{n+1/a-1} = 1 - \frac{1/a-1}{n+1/a-1}$? Thank you, I appreciate your help. –  Jamil_V Sep 11 '13 at 4:50
Nevermind, I see it now. Thank you very much. –  Jamil_V Sep 11 '13 at 4:56
add comment

Check this and then use the squeeze theorem with:

$$(1)\;\;a>1\implies 1\le\sqrt[n] a\le\sqrt[n]n\;\;,\;\;\text{for}\;\;n\ge a\\(2)\;\;a<1\implies\;\text{use arithmetic of limits with}\;\;b:=\frac1a>1\ldots$$

share|improve this answer
Thank you very much for showing me that link. It was very helpful. –  Jamil_V Sep 9 '13 at 4:12
add comment

Case 1: Suppose that $a\gt 1$. Let $a=1+\delta$, where $\delta$ is positive.

We claim that if $n \ge 1$ then $$1\le a^{1/n}\lt 1+\frac{\delta}{n}.\tag{1}$$ The fact that $\lim_{n\to\infty} a^{1/n}=1$ is an immediate consequence of Inequality (1).

To prove (1), suppose to the contrary that $a^{1/n}\gt 1+\frac{\delta}{n}$. Then $$a=(a^{1/n})^n\gt \left(1+\frac{\delta}{n}\right)^n.\tag{2}$$ But by the Bernoulli Inequality, $$\left(1+\frac{\delta}{n}\right)^n \ge 1+n\frac{\delta}{n}=1+\delta.\tag{3}$$ From (2) and (3) we conclude that $a\gt 1+\delta$, contradicting the fact that $a=1+\delta$.

Case 2: Suppose that $0\lt a\lt 1$. Let $b=\frac{1}{a}$. Then $b\gt 1$. By Case 1, $b^{1/n}$ has limit $1$, and therefore so does $a^{1/n}$.

Remarks: $1.$ The Bernoulli Inequality states that if $t\gt -1$, then $(1+t)^n\ge 1+nt$. We only need it for $t\gt 0$. In that case, it is an immediate consequence of the Binomial Theorem. There is also a quite simple induction proof.

$2.$ We gave a quite formal proof. But it comes down to the fact that a number $\gt 1$ raised to a large enough power is very large, and in particular greater than $a$.

share|improve this answer
I am somewhat familiar with Bernoulli's Inequality; there is a small section at the end of this chapter in my textbook about it. This post helps to better understand how it can be used. Thank you very much. –  Jamil_V Sep 9 '13 at 4:15
You are welcome. The Bernoulli Inequality can be useful. –  André Nicolas Sep 9 '13 at 4:22
This proves it for $a > 1$. The OP also asked for $0 < a < 1$. –  marty cohen Sep 11 '13 at 4:34
I did not notice. Will try to deal with that tomorrow. It is the same, instead of calling $a$ by the name $1+\delta$, one calls it $\frac{1}{1+\delta}$. –  André Nicolas Sep 11 '13 at 4:50
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.