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

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

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

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## marked as duplicate by Martin Sleziak, draks ..., LeGrandDODOM, Ivo Terek, Mark BennetJun 19 '15 at 15:37

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
– Martin Sleziak Jun 19 '15 at 12:51

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):

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

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

$$\sqrt[n]n\xrightarrow[n\to\infty]{}1$$

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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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The AM-GM inequality is enough. Since we have the following telescopic product: $$n= 1\cdot\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\tag{1}$$ we have: $$1 = \sqrt[n]{1} \leq \sqrt[n]{n} \leq \frac{n+H_{n-1}}{n} \leq 1+\frac{\log (2n-1)}{n}\tag{2}$$ from: $$H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}\leq \sum_{k=1}^{n-1}\log\left(\frac{2k+1}{2k-1}\right)=\log(2n-1).\tag{3}$$

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