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I had a test today and there was an extra problem I couldn’t solve.

Put the $sqrt[2]{2}, sqrt[3]{3}, dots, sqrt[100]{100}$ numbers in to an increasing order.

I just have no idea. I can handle the numbers $sqrt[2^k]{2^k}$ numbers, but I can’t solve the problem.

Consider the function $f(x)=x^{1/x}$ defined for $x>0$.

Then $f(x)=exp(frac{log x}{x})$ and so
$$
f'(x)=f(x)frac{1-log x}{x^2}
$$
which is positive over $(0,e)$ and negative over $(e,infty)$. So we know that
$$
sqrt[n]{n}>sqrt[n+1]{n+1}
$$
for $nge 3$.

Also $(sqrt{2})^6=8$, $(sqrt[3]{3})^6=9$, hence $sqrt{2}<sqrt[3]{3}$. Moreover $sqrt{2}=sqrt[4]{4}$.

Hint only.

I suppose you know how to study functions, you know power functions and also that you know how to de the derivative a combination of functions.

Then, can you find a function that goes through all the points you are describing?

Hint 1 If you have differential calculus available: Differentiate $x mapsto x^{1 / x}$ (or, since $log$ is increasing, $x mapsto log(x^{1 / x}) = frac{log x}{x}$) to find a single maximum that lies between $2$ and $3$.

Then, it remains only to find where in the list $sqrt{2}$ fits, but you've already written that you can handle the cases $sqrt[2^k]{2^k}$.

Hint 2 Alternatively, we can compare the ratio $frac{sqrt[n + 1]{n + 1}}{sqrt[n]{n}}$ of successive values; raising that ratio to the $n(n + 1)$ power gives $$frac{(n + 1)^n}{n^{n + 1}} = frac{1}{n} left(1 + frac{1}{n}right)^n .$$ But $left(1 + frac{1}{n}right)^n$ increases monotonically to $e < 3$.

Note that $ln n^{1/n}=frac {ln n}{n}$

Upon differentiating the function $$f(x)=frac {ln x}{x}$$
we get $$f'(x)=frac {1-ln x}{x^2}<0$$ for $xge 3$

Thus the sequence $n^{1/n}$ is decreasing for $nge 3$

The ordering is $$ 1<2^{1/2}<3^{1/3}$$ and $3^{1/3}>4^{1/4}>5^{1/5}>.....$