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With $a_{1}=1, a_{n}=n/(n-1),\text{ when }n > 1$ apply $c_{n}=\sqrt[n]{a_{1}\cdots a_{n}}$ which implies $\ln(c_{n})=\sum\ln(a_{n})/n$ to prove that $\sqrt[n]{n} \rightarrow 1$

So I know the easier way to get $\sqrt[n]{n} \rightarrow 1$, but I have to use apply $c_{n}$

So $\ln(c_{n})=\ln\left(\frac{(\frac{n}{n-1})}{n}\right)$ $\rightarrow$ $\frac{\ln(n)}{n} - \frac{\ln(n-1)}{n}$ $\rightarrow \ln(\sqrt[n]{n})-\ln(\sqrt[n]{n-1})$

Its at this point that I'm not sure what to continue doing. I could raise everything by e and then take the limit...

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You've got a mess there, with $\sum \ln(a_n)/n$ where it really should be $\sum_{i} \ln(a_i)/n$ –  Thomas Andrews Sep 24 '12 at 18:23
    
Ah, sorry but I hope that slight typo doesn't change the substance of the post. –  Edgar Aroutiounian Sep 24 '12 at 18:24
    
The title: "Using logs to show $\,\sqrt[n] n\,$"...to show what? –  DonAntonio Sep 24 '12 at 18:39
    
I'm terribly sorry, I had it as goes to 1. Not sure where it went.edit, fixed. –  Edgar Aroutiounian Sep 24 '12 at 18:41
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4 Answers

up vote 1 down vote accepted

I think you're expected to write: $$ \ln(c_n)=\ln (a_1a_2\cdots a_n)^{1/n}={1\over n}\sum_{i=2}^n \ln\underbrace{{i\over i-1}}_{a_i}={1\over n}\underbrace{\sum_{i=2}^n \bigl(\ln i-\ln(i-1)\bigr)}_{\text{telescoping sum}}={1\over n}(\ln n){\buildrel n\rightarrow \infty\over\longrightarrow}0.$$ So $c_n=e^{\ln c_n}\rightarrow e^0=1$.

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@Edgar Aroutiounian Of course, you'd need to know $\ln n\over n$ tends to zero as $n$ tends to $\infty$. And from your comment in another answer, to be able to this without using L'Hopital... –  David Mitra Sep 24 '12 at 19:15
    
You don't need L'Hôpital's rule. You can use the fact that $e^x > \dfrac{x^2}{2!}$. –  Ayman Hourieh Sep 24 '12 at 20:07
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I think another way would be

$\lim\limits_{n\to\infty}\sqrt[n]{n}=\lim\limits_{n\to\infty}\mathbb{e}^{\ln\left(\sqrt[n]{n}\right)}=\lim\limits_{n\to\infty}\mathbb{e}^{\frac{1}{n}\ln(n)}=\mathbb{e}^{\lim\limits_{n\to\infty}\frac{\ln(n)}{n}}\overset{Del'Hospital}{=}\mathbb{e}^{\lim\limits_{n\to\infty}\frac{\ln(n)'}{n'}}=\mathbb{e}^{\lim\limits_{n\to\infty}\frac{1}{n}}=\mathbb{e}^{0}=1$.

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thank you, but I'm not sure I can get away with l'hospital's rule just yet. –  Edgar Aroutiounian Sep 24 '12 at 19:12
    
Even without the l'hospital rule, the limit $\lim_{n \to \infty} \frac{ \log (n) }{n} = 0 $ is "standard". –  Andy Sep 24 '12 at 21:58
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Since $\,n\geq 1\Longrightarrow \sqrt[n] n\geq 1\,\,,\,\,\forall\,n\in\Bbb N\,$ , so we can write $\,\sqrt[n] n=1+c_n\,\,,\,\,c_n\geq 0\,\,\,\forall\,n\in\Bbb N,\,$ , and then:

$$n=(1+c_n)^n=\sum_{k=0}^n\binom{n}{k}c_n^k\geq\frac{n(n-1)}{2}c_n^2\Longrightarrow$$

$$\Longrightarrow c_n\leq\sqrt\frac{2}{n-1}$$

so finally, using the squeeze theorem:

$$1\leq\sqrt[n] n\leq 1+\sqrt\frac{2}{n-1}\xrightarrow [n\to \infty]{} 1$$

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Let $L = \lim\limits_{n\to\infty}\sqrt[n]{n}$. Thus, $\log(L) = \lim\limits_{n\to\infty}\frac{1}{n} \log(n)$. Since logarithmic growth is slower than linear growth, we have $$\log(L) = \lim\limits_{n\to\infty}\frac{1}{n} \log(n) = 0$$ Thus, $L = 1$.

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