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My solution like this:

Let $\sqrt[n]n-1=\alpha_n$

  $\sqrt[n]n\geq1$  so $\alpha_n\geq 0$

For $n\geq2 $

$$n=(1+\alpha_n)^n=1+n\cdot \alpha_n+\frac{n\cdot (n-1)}{2}\cdot\alpha _n^2+\ldots+\alpha _n^n\geq \frac{n\cdot (n-1)}{2}\cdot\alpha _n^2$$

For $n\geq2 $

 $n-1\geq \frac{n}{2}$  then $n\geq \dfrac{n^2\cdot\alpha _n^2}{4}$ so that $0\le \alpha_n\leq \dfrac{2}{\sqrt n}$  and $$\lim\limits_{n\to \infty}\alpha_n=0$$

Finally $$\lim\limits_{n\to \infty}\sqrt[n]{n}=\lim\limits_{n\to \infty}(1+\alpha_n)=1$$

Are there any other solutions?

TShiong
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deepblue
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    A little too tired to check your work on this, but a quick alternative solution is to use L'Hopital's rule to show that $\log(n^{1/n})=\frac{\log(n)}{n}$ converges to $0$, hence the original sequence converges to $e^0=1$. – M W Aug 18 '23 at 11:43
  • @MW: that would be a circular argument. No need to define exponential function and then logarithmic function at this point. – Mittens Aug 18 '23 at 13:51
  • Thank you Mittens. – deepblue Aug 18 '23 at 14:14
  • The notation $\mathbb N^+$ is redundant as all the elements of $\mathbb N$ are positive. – K.defaoite Aug 18 '23 at 16:15

1 Answers1

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By Taylor's theorem, $$n^{1/n}=\exp\left(\frac{1}{n}\ln n\right)=1+\frac{\ln n}{n}+O\left(1/n^2\right)$$ as $n\to\infty$. The limit is $1$.

bob
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  • But $n \in \mathbb{N^+}$. İn your solutions $n \in \mathbb{R^+}$ Can you use derivative for natural numbers? – deepblue Aug 18 '23 at 12:04
  • The Maclaurin series for $e^x$ is defined for all $x\in\mathbb{C}$, which also includes natural numbers. So as far as I can tell, there should be no problem. – bob Aug 18 '23 at 12:10
  • Thank you, but is there a solution in $\mathbb{N^+}$ – deepblue Aug 18 '23 at 12:21