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Possible Duplicate:
$\lim_{n \to +\infty} n^{\frac{1}{n}} $

I know that

$$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$

and I can imagine that $n$ grows linearly while $n$th root compresses it exponentially and therefore the result is $1$, but how do I calculate it?

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    $\begingroup$ Look at the limit of the logarithm instead: $\lim\limits_{n\to\infty}\frac1n\ln n$. What’s $\lim\limits_{x\to\infty}\frac{\ln x}x$? $\endgroup$ Commented Jun 5, 2012 at 9:44
  • $\begingroup$ So the "trick" is $$a^b=e^{b*\ln{a}}$$ $\endgroup$
    – Santhos
    Commented Jun 5, 2012 at 10:28

6 Answers 6

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Perhaps one of the most elementary ways to prove it: since $\,n\geq 1\,\,\forall n\in\mathbb{N}\,$ , we can put$$\sqrt[n]{n}=1+c_n\,,\,c_n\geq0\Longrightarrow n=(1+c_n)^n\geq \frac{n(n-1)}{2}c_n^2$$using the binomial expansion ,so that$$0<c_n\leq\sqrt\frac{2}{n-1}$$and now just apply the squeeze theorem and get $\,c_n\to 0\,$, which is precisely what we need.

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  • $\begingroup$ what is this cn?? $\endgroup$
    – S L
    Commented Jun 5, 2012 at 17:28
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    $\begingroup$ It's a number $>0$ that depends on $n$. $\endgroup$ Commented Jun 5, 2012 at 19:30
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    $\begingroup$ @experimentX: Specifically, $c_n$ by definition is $\sqrt[n]{n}-1$. It is a case of introducing notation to make the exposition easier. $\endgroup$ Commented Jun 6, 2012 at 1:41
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$$\lim_{n\rightarrow\infty}\sqrt[n]n$$ $$=\lim_{n\rightarrow\infty}e^{\frac{\ln(n)}{n}}$$ and as we know that $\lim_{n\rightarrow\infty}\frac{\ln(n)}{n} = 0$ {apply l'Hospital's rule}
So $$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$

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Claim: For each $a>1$, there exists $N$ such that $n<a^n$ for all $n>N$.

Proof: Write $a=1+b$. By the binomial theorem, $a^n=(1+b)^n\geq \frac{1}{2}n(n-1)b^2$ when $n\geq 2$. Thus $\frac{a^n}{n} \geq \frac{1}{2}(n-1)b^2$. It follows that if $N$ is at least $2/b^2+1$, then when $n>N$, $n<a^n$.


As a consequence, for each $a>1$, there exists $N$ such that $1\leq n^{1/n}<a$ for all $n>N$, and this implies that $\lim\limits_{n\to\infty}n^{1/n}=1$.

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    $\begingroup$ DonAntonio's answer is simpler. $\endgroup$ Commented Jun 5, 2012 at 10:54
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    $\begingroup$ Yes, thanks for noticing it. One of the things I've never forgotten since my late prof. in $1$st undergraduate year taught us this proof: I think I drooled for a while completely fascinated for the simplicity of the proof, compared with what several books showed. $\endgroup$
    – DonAntonio
    Commented Jun 5, 2012 at 11:00
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    $\begingroup$ @DonAntonio: Inspired by your post I cleaned mine up a little (but didn't essentially change the method). $\endgroup$ Commented Jun 5, 2012 at 11:01
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Let's see a very elementary proof. Without loss of generality we proceed replacing $n$ by $2^n$ and get that: $$ 1\leq\lim_{n\rightarrow\infty} n^{\frac{1}{n}}=\lim_{n\rightarrow\infty} {2^n}^{\frac{1}{{2}^{n}}}=\lim_{n\rightarrow\infty} {2}^{\frac{n}{{2}^{n}}}\leq\lim_{n\rightarrow\infty} {2}^{\frac{n}{\dbinom{n}{2}}}=2^0=1$$

By Squeeze Theorem the proof is complete.

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    $\begingroup$ what did you do near the last step?? $\endgroup$
    – S L
    Commented Jun 5, 2012 at 17:27
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    $\begingroup$ @experimentX: consider the fact that $2^n$ is the sum of binomial coefficients for given n. Then you take all all the terms away except that one and get an useful inequality. See here: proofwiki.org/wiki/Sum_of_Binomial_Coefficients_for_Given_n $\endgroup$ Commented Jun 5, 2012 at 17:32
  • $\begingroup$ I understand that ... thanks !! $\endgroup$
    – S L
    Commented Jun 5, 2012 at 17:34
  • $\begingroup$ @experimentX: you're welcome! $\endgroup$ Commented Jun 5, 2012 at 17:36
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One could use the fact that for a sequence of positive terms, if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} \root n\of {a_n}$ and the two limits are equal. A proof of this general fact can be found in these notes of Pete L. Clark. This result can also be found in many analysis texts; e.g., baby Rudin.

That your sequence has limit $1$ is easily shown using the above fact. A detailed proof that your sequence has limit $1$, based on the proof of the above fact, can be found in this thread.

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  • $\begingroup$ See also here and here for the result on relation between $\frac{a_{n+1}}{a_n}$ and $\sqrt[n]{a_n}$. $\endgroup$ Commented Jun 5, 2012 at 16:41
  • $\begingroup$ The linked text later became a part of the notes on Honors Calculus (Wayback Machine). $\endgroup$ Commented Jul 5, 2022 at 5:37
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Let$$y=n^{\frac{1}{n}}$$ $$\log y=\log( n^{\frac{1}{n}})$$ $$\Rightarrow \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{\log n}{n}$$ We have $\left(\frac{\infty}{\infty}\right)$ form so we apply L'hospital rule, we get, $$ \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{\frac{1}{n}}{1}$$ $$\therefore \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{1}{n}=0$$ Hence $$\lim_{n\to\infty} y=e^{0}=1.$$ It follows that, $$\lim_{n\to\infty}n^{\frac{1}{n}}=1.$$

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