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$\lim_{n \to +\infty} n^{\frac{1}{n}} $
Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$

Find the limit of $u_n=\sqrt[n]{n}$ I try to prove $\sqrt[n]{1}<\sqrt[n]{n}<$something that some thing <

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marked as duplicate by DonAntonio, Douglas S. Stones, Davide Giraudo, Nameless, Martin Argerami Jan 5 '13 at 12:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Read… – DonAntonio Jan 5 '13 at 11:04
up vote 1 down vote accepted

This is a classical sequence. It's an example in any instructional first course of analysis. There are several ways to obtain the limit of this sequence. But first of all we must prove that this sequence converge.

For all $\epsilon>0$ exist $n_\epsilon\in\mathbb{N}$ such that $n>n_\epsilon$ implies $n\cdot \epsilon< e^{n\cdot \epsilon+1}$. \begin{align} n\cdot \epsilon < e^{n\cdot \epsilon+1}\implies & \ln (n\cdot \epsilon) < n\cdot \epsilon+1 \\ \implies & \frac{1}{n}\ln (n\cdot \epsilon) < \frac{1}{n}(n\cdot \epsilon+1) \\ \implies & \ln\big(\sqrt[n]{n}\cdot \sqrt[n]{\epsilon}\big) < 1\cdot \epsilon+\frac{1}{n} \\ \implies & \ln\sqrt[n]{n}+ \ln\sqrt[n]{\epsilon} < 1\cdot \epsilon+\frac{1}{n} \\ \implies & \ln\sqrt[n]{n} < 1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon} \\ \implies & \sqrt[n]{n} < e^{(1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon})} \\ \end{align} Then $$ 1< \sqrt[n]{n} < e^{(1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon})} $$ And $\displaystyle\lim_{n\to\infty}\Big(1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon}\Big)=\epsilon$ implies that for all $\epsilon>0$ exist $n_\epsilon\in\mathbb{N}$ such that $n>n_\epsilon$ we have $$ 1< \sqrt[n]{n} < e^{ \epsilon}. $$ Then we have the result: $\lim_{n\to 0}{\sqrt[n]{n}}=1$

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Since $\sqrt[n]{n} > 1$, write $\sqrt[n]{n} = 1 + a_n$ where $a_n > 0$. Thus: $$ n = (1 + a_n)^n $$

Apply the binomial theorem to the RHS. Can you use the above to squeeze $a_n$ and prove it converges?

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I have used this but i only prove that : $$a_n \leq \frac{n-1}{n}$$ something wrong – Haruboy15 Jan 5 '13 at 10:36
Pick another term of the expansion? It also looks like you made a mistake in expanding $(1+a_n)^n$. Can you show us your work? – Ayman Hourieh Jan 5 '13 at 10:41
@AymanHourieh - A very nice solution! I've known this problem for a long time but never thought about the binomial theorem. – Ofir Jan 5 '13 at 11:46
@Ofir I don't take credit for it. It's from Rudin's Principles. :) – Ayman Hourieh Jan 5 '13 at 12:16

Its a lot simpler actually. Since $log(x)$ is a continuous function you have that $a_n\rightarrow a$ implies $log(a_n)\rightarrow log (a)$ whenever logs are defined. Try using this.

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can you show me more; I have just studied this – Haruboy15 Jan 5 '13 at 10:36
He means: Try to find the limit of $\log(\sqrt[n]{n})$. – barto Jan 5 '13 at 10:41

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