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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.

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There was a discussion about this yesterday here:… – Abhimanyu Arora Feb 17 '14 at 9:26
your solution does not have errors. – Babak Miraftab Feb 17 '14 at 9:26
You must assume that $a>0$. – gammatester Feb 17 '14 at 9:28
up vote 4 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 $$ \lim_{n\to\infty}\left(\sqrt[n]{n}-1\right)=\lim_{n\to\infty}a_{n}=0\iff\lim_{n\to\infty}\sqrt[n]{n}=1 $$

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This seems extremely complicated - what are the advantages versus the sort of solution I proposed in my answer? – Jack M Feb 17 '14 at 21:41
Advantages? I don't know. I think it's a nice proof. – Amihai Zivan Feb 17 '14 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 '14 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 $\mathbb R$.

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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 '14 at 5:48

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