# Largest element in the set $\{1,\sqrt{2},\sqrt[3]{3},…,\sqrt[n]{n}\}$ [closed]

What is the largest element in the set $$\;\left\{1,\sqrt{2},\sqrt[3]{3},...,\sqrt[n]{n}\right\}$$?

Once I write down the numbers, it seems like the largest element will be $$\sqrt[3]{3}$$, but I couldn't come up with an explicit proof.

It looks like the sequence $$a_n=\sqrt[n]{n}$$ increases up to $$n=3$$ and then decreases down to $$1.$$

• One approach using calculus is to study the function $f(x) = x^{1/x} = e^{\log(x)/x}$. You can find where it's increasing, where it's decreasing and where it has it's maximum. Then you will be left with just checking the integers nearest to the maximum. – Winther Dec 22 '18 at 23:19
• Thanks alot Winther. I posted an answer following your hint – DD90 Dec 22 '18 at 23:43
• For subsequent questions, it would be best to provide more context (see the above explanation). But here is a proof that may be more instructive than those you have gotten so far: $n \le \sqrt{2}^n$ for every natural $n \ge 4$ (which you can prove by induction). Thus $\sqrt[n]{n} = \sqrt{2}$ for every natural $n \ge 4$. Then compare those $n < 4$. – user21820 Dec 23 '18 at 6:54

It's possible to answer this without calculus as well. Consider two consecutive terms $$\sqrt[n]n$$ and $$\sqrt[n+1]{n+1}$$, and compare them: $$\sqrt[n]n\,?\,\sqrt[n+1]{n+1}\\ n^{n+1}\,?\,(n+1)^n\\ n\,?\,\left(\frac{n+1}{n}\right)^n\\ n\,?\,\left(1+\frac1n\right)^n$$ where the right-hand side is well-known to never exceed $$e$$ (and even without that, showing that it never exceeds $$3$$ is not too difficult). So clearly, for $$n\geq3$$, the question mark will be a $$\geq$$, meaning all terms after $$\sqrt[3]3$$ will be smaller than $$\sqrt[3]3$$.
By the comment and the answer hint which was given I managed to setup the solution. Observe that $$f'(x)=\frac{f(x)(1-\ln(x)}{x^2}$$ So the function increases upto $$e$$ and then decreases from there onwards.
And we can notice that $$f(2)=f(4)$$ and also both 3 and 4 lie on the decreasing side. Thus $$f(3)>f(4)$$.
Hence $$f(3)$$ should be the maximum value.
• It would be preferable to also add the derivation (how you found the maximum was $e$). This makes the answer more useful to others. – Winther Dec 22 '18 at 23:45
The function $$f(x)=x^{1/x}$$ is increasing wherever $$g(x)=\log f(x)$$ is increasing. Now $$g(x)=\frac{\log x}{x}$$ and $$g'(x)=\frac{1-\log x}{x^2}$$ Can you finish?