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Say I need to compute $\lim_{n \to \infty} \sqrt[n]{\sqrt[n]{n}}$ and write $$\lim_{n \to \infty} \sqrt[n]{\sqrt[n]{n}} = \lim_{n \to \infty} \sqrt[n]{1} = 1$$

by the fact that $\lim_{n \to \infty} \sqrt[n]{n} = 1$ (i.e., assume that this fact fact is something that can be used without proof).

Question: how bad (or good) is this solution?

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    $\begingroup$ That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=\sqrt{(n^2)^{1/n^2}} \rightarrow 1$ because of the limit that you may use. $\endgroup$
    – Aphelli
    Jan 21, 2019 at 17:33

5 Answers 5

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The solution is not ok that way.

Your argument would fail, for example, for $\left( 1+\frac{1}{n}\right)^n$.

But you can still use the fact $\lim_{n \to \infty} \sqrt[n]{n} = 1$ as follows: $$1 \leq \sqrt[n]{n} <2 \mbox{ for } n \mbox{ large enough }$$ $$\Rightarrow 1 \leq \sqrt[n]{\sqrt[n]{n}} < \sqrt[n]{2} \stackrel{n \to \infty}{\longrightarrow}1$$

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$$n^{\frac{1}{n^2}} = e^{\frac{1}{n^2}\ln n}$$ Then use l'Hopital: $$\lim_{n\rightarrow\infty}{n^{\frac{1}{n^2}} } = e^{\lim_{n\rightarrow\infty}{\frac{\ln n}{n^2}}} = e^{\lim_{n\rightarrow\infty}{\frac{1}{2n^2}}} = 1$$ Note that I am allowed to move the limit inside the exponent because of the continuity.

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Bad: you replace part of the expression with its limit while leaving the rest untouched, which is strictly forbidden.

The faster way consists in calculating the limit of the log, using its asymptotic properties: $$\log\Bigl(n^{\tfrac1{n^2}}\Bigr)=\frac{\log n}{n^2}\to 0\enspace\text{ as }n\to\infty,$$ so by continuity of the inverse function, $$ \lim_{n\to\infty}n^{\tfrac1{n^2}} =\mathrm e^0=1.$$ The same proof is valid for any exponent $\alpha>0$: $$ \lim_{n\to\infty}n^{\tfrac1{n^\alpha}}=\mathrm e^0=1.$$

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$n>1;$

$1 \lt n^{1/n^2} \lt (n^2)^{1/n^2}.$

Take the limit.

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Let $x$ be real and non negative such that $n^{\frac{1}{n^2}}=1+x$. Then $n=(1+x)^{n^2}$, so by Bernoulli's inequality we have $n\ge1+n^2x$ or $x\le\frac{1}{n}-\frac{1}{n^2}$. Take the limit and you get that $x\le0$. I leave it to you to show is that the root can't actually be strictly smaller than $1$.

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