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.

Hi guys this was a practice problem I was given, can anyone help me out on it? This is the problem: Show that $\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{a} =1$ and the following is what I came up with, any help is appreciated, thank you!

$\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{a}$

= $\displaystyle \lim_{n \rightarrow \infty} e^{ln(\sqrt[n]{a})}$

= $e^{\lim_{n \rightarrow \infty}(\frac{1}{n}) ln(a)}$

= $e^{\lim_{n \rightarrow \infty}(0 \cdot ln(a))}$

= $e^0$

= 1.

share|improve this question
    
There was a discussion about this yesterday here: math.stackexchange.com/questions/678309/… –  Abhimanyu Arora Feb 17 at 9:26
4  
your solution does not have errors. –  Babak Miraftab Feb 17 at 9:26
2  
You must assume that $a>0$. –  gammatester Feb 17 at 9:28

2 Answers 2

up vote 3 down vote accepted

I think that this one is due to Courant:

For all $n\in\mathbb{N}$ define $a_{n}$ such that $$ a_{n}=\sqrt[n]{n}-1 $$ Note that for every $n\in\mathbb{N}$, $a_{n}\geqslant0$. Now rewrite as: \begin{eqnarray*} n & = & \left(a_{n}+1\right)^{n}\\ & = & \sum_{k=0}^{n}\binom{n}{k}a_{n}^{k}\\ & \geqslant & \binom{n}{2}a_{n}^{2} \end{eqnarray*} We get $n\geqslant\frac{n(n-1)}{2}a_{n}^{2}$.

After rearranging:

$$ a_{n}\leqslant\sqrt{\frac{2}{n-1}}\xrightarrow[n\to\infty]{}0 $$ Concluding that $$ \lim_{n\to\infty}\left(\sqrt[n]{n}-1\right)=\lim_{n\to\infty}a_{n}=0\iff\lim_{n\to\infty}\sqrt[n]{n}=1 $$

share|improve this answer
    
This seems extremely complicated - what are the advantages versus the sort of solution I proposed in my answer? –  Jack M Feb 17 at 21:41
1  
Advantages? I don't know. I think it's a nice proof. –  Amihai Zivan Feb 17 at 21:44
    
Its bit surprising that you have solved a different question concerning $\lim_{n \to \infty}\sqrt[n]{n}$ whereas OP asked for $\lim_{n \to \infty}\sqrt[n]{a}$ for $a > 0$. Even more difficult is to be believe that OP accepted your answer. –  Paramanand Singh Feb 22 at 5:38

Take any number $r>1$. Eventually $r^n>a$ and so eventually $\sqrt[n]a<r$. Now take $r<1$. Eventually $r^n<a$ and so eventually $\sqrt[n]a>r$.

Calculating $\lim\ r^n$ rigorously is a little tricky and depends on how deep you want to go, see this answer for one approach.

I did also use the fact that the $n$-th root function is increasing, but I'm not sure how to prove that without diving into the definition of the ordering in $\mathbb R$.

share|improve this answer
    
Your approach is very simple so +1 for that. Regarding the increasing nature of $n$-th root its not difficult at all. If $a > b > 0$ and $n$ is positive integer then it is very easy to show that $a^{n} > b^{n}$ (just multiply $a > b\,\,$ $n$ times). Next let $c > d > 0$. Clearly if $c^{1/n} \leq d^{1/n}$ then from the previous result we must have $c \leq d$ which is not the case. Hence $c^{1/n} > d^{1/n}$. –  Paramanand Singh Feb 22 at 5:48

Your Answer

 
discard

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.