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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 $1.$

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    $\begingroup$ 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. $\endgroup$ – Winther Dec 22 '18 at 23:19
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    $\begingroup$ Thanks alot Winther. I posted an answer following your hint $\endgroup$ – DD90 Dec 22 '18 at 23:43
  • $\begingroup$ 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 $n < 4$. $\endgroup$ – user21820 Dec 23 '18 at 6:54
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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 $\sqrt[3]3$.

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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.

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    $\begingroup$ It would be preferable to also add the derivation (how you found the maximum was $e$). This makes the answer more useful to others. $\endgroup$ – Winther Dec 22 '18 at 23:45
  • $\begingroup$ Thanks and I edited $\endgroup$ – DD90 Dec 22 '18 at 23:53
  • $\begingroup$ One parenthesis is missing $\endgroup$ – Pedro A Dec 23 '18 at 0:26
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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?

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