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This was a practice problem I was given.

Show that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$.

My argument is as follows: \begin{align} \lim_{n \rightarrow \infty} \sqrt[n]{a} &= \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.\end{align}

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  • $\begingroup$ There was a discussion about this yesterday here: math.stackexchange.com/questions/678309/… $\endgroup$ – Abhimanyu Arora Feb 17 '14 at 9:26
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    $\begingroup$ your solution does not have errors. $\endgroup$ – Babak Miraftab Feb 17 '14 at 9:26
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    $\begingroup$ You must assume that $a>0$. $\endgroup$ – gammatester Feb 17 '14 at 9:28
  • $\begingroup$ You should simply consider rewriting $\sqrt[n]{a}$ as $a^{\frac{1}{n}}$. As $n \to \infty$, $\frac{1}{n} \to 0$ $\endgroup$ – Paras Khosla Feb 5 at 15:12
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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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  • $\begingroup$ This seems extremely complicated - what are the advantages versus the sort of solution I proposed in my answer? $\endgroup$ – Jack M Feb 17 '14 at 21:41
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    $\begingroup$ Advantages? I don't know. I think it's a nice proof. $\endgroup$ – Amihai Zivan Feb 17 '14 at 21:44
  • $\begingroup$ 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. $\endgroup$ – Paramanand Singh Feb 22 '14 at 5:38
  • $\begingroup$ @ParamanandSingh: it seems customary that POs accept answers without checking them... $\endgroup$ – Yves Daoust Sep 9 at 11:52
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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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  • $\begingroup$ 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 $c^{1/n} > d^{1/n}$. $\endgroup$ – Paramanand Singh Feb 22 '14 at 5:48
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There is an essential piece in your reasoning, as you swap a limit and a function (the logarithm): you must state that the function is continuous.

You may have simply written $$\log\left(\lim_{n\to\infty}\sqrt[n]a\right)=\lim_{n\to\infty}\log\left(\sqrt[n]a\right)=\lim_{n\to\infty}\left(\frac{\log a}n\right)=0.$$

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