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I was trying to rigorously show that $\lim \limits_{n\to \infty} \sqrt[n]{\frac 1n} = 1$ but so far I have not been very successful. I simply built my argument on what has been already shown, namely that $\lim \limits_{n\to \infty} \sqrt[n]{p}=1$ for any $p>0$ - the limit then just becomes a special case of this. Do you know of any other way of justifying this?

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  • $\begingroup$ Wouldn't that be $0$??? I mean, $\lim \limits_{x\to 0} \sqrt[n]{\frac 1n}$ is "hard", and $\lim \limits_{x\to \infty} \left(\frac 1n\right)^n$ is "hard", but this one looks rather straightforward. $\endgroup$ Jan 3 '16 at 17:50
  • $\begingroup$ See my answer here for a more elementary proof that $\lim_{n\to\infty}\sqrt[n]n=1$. It's not difficult to apply that proof to this problem. $\endgroup$
    – Arthur
    Jan 3 '16 at 18:03
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$$ n^{-1/n}=e^{-\frac{1}{n}\log n} $$ and we know $\frac{\log n}{n}$ tends to $0$, so...

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Notice, let $\frac 1n=t\implies t\to 0$ as $n\to \infty$, $$\lim_{n\to \infty}\sqrt[n]{\frac{1}{n}}=\lim_{n\to \infty}\left(\frac 1n\right)^{\frac 1n}$$ $$=\lim_{t\to 0}t^{t}$$ $$=\exp\lim_{t\to 0}\ln(t^t)$$ $$=\exp\lim_{t\to 0}(t\ln t)$$ $$=\exp\lim_{t\to 0}\frac{\ln t}{1/t}$$

Using L'Hospital's rule for $\frac{\infty}{\infty}$ form $$=\exp\lim_{t\to 0}\frac{\frac 1t}{-1/t^2}$$ $$=\exp\lim_{t\to 0}(-t)=e^0=\color{red}{1}$$

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Let $$\sqrt[n]{\frac{1}{n}}=\frac{1}{1+t_n}.\tag{1}$$ We want to show that $\lim_{n\to\infty}t_n=0$. It is clear that if $n\gt 1$ then $t_n\gt 0$.

From (1) and the binomial theorem we have for $n\gt 1$ $$n=(1+t_n)^n\ge 1+nt_n +\frac{n(n-1)}{2}t_n^2\gt \frac{n(n-1)}{2}t_n^2.$$ Thus $$0\lt t_n \lt \sqrt[2]{\frac{2}{n-1}},$$ and the rest follows by squeezing.

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