# Limit of an $n$-th Root Proof

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.

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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

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 $\epsilon > 0$.

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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

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 $a$.

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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

Check this and then use the squeeze theorem with:

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

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Thank you very much for showing me that link. It was very helpful. – Jamil_V Sep 9 '13 at 4:12