Put the $\sqrt[n]{n}$ numbers in to an increasing order

Question

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.

Answer 1

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 $n\ge 3$.

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

Answer 2

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?

Answer 3

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 $e < 3$.

Answer 4

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 $x\ge 3$

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

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