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Find: $$\lim_{n\to\infty} \sqrt[n]{n}$$

How does one not having any intuition able to do this?

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Have you had l'Hopital's Rule in your course yet? You can express your problem as $lim x \rightarrow \infty (x^{1/x})$ and treat this as an "indeterminate powers" limit. –  RecklessReckoner Apr 15 '13 at 1:20

3 Answers 3

up vote 12 down vote accepted

Note that $n^{\frac{1}{n}} = e^{\frac{1}{n}\log(n)}$. Since $\frac{1}{n}\log(n)\to 0$ as $n\to\infty$ and since the exponential function is continuous, we have $\lim_{n\to\infty} n^{\frac{1}{n}} = \lim_{n\to\infty} e^{\frac{1}{n}\log(n)} = e^0 = 1$.

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Hints (for you to work out and justify):

$$\forall\,n\in\Bbb N\;,\;\;\sqrt[n]n\ge 1\implies \sqrt[n]n=1+c_n\;,\;\;c_n\ge 0\implies$$

$$n=(1+c_n)^n=\sum_{k=0}^n\binom{n}{k}c_n^k\ge \frac{n(n-1)}{2}c_n^2\implies c_n\le \sqrt\frac{2}{n-1}\implies c_n\xrightarrow [n\to\infty]{}0\implies$$


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Step One: $1^1=1$, $2^\frac{1}{2}>1$, suppose $n>1 \implies n^\frac{1}{n}>1^\frac{1}{n}=1$.

Step Two: Define $\alpha_n=n^\frac{1}{n}-1$. By the Binomial Formula for every index $n>2$, $n=(1+\alpha_n)^n=1+n\alpha_n+\frac{n(n-1)}{2}\alpha_n^2+\dots>1+\frac{n(n-1)}{2}\alpha_n^2$, since $\alpha_n>0$. $n>1+\frac{n(n-1)}{2}\alpha_n^2\implies 1>\frac{n}{2}\alpha_n^2\implies \frac{2}{n}>\alpha_n^2$ and by the comparison lemma and since the product of convergent sequences converges to the product of the limits, you can see that $\lim_{n \to \infty} \alpha_n=0$.

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