Finding the supremum of $\left\{n^{\frac{1}{n}}\;\middle\vert\;n\in\mathbb{N}\right\}$

Question

$\left\{n^{\frac{1}{n}}\;\middle\vert\;n\in\mathbb{N}\right\}$

What is the Supremum of the above set?

I consider the function $f(x)= x^{\frac{1}{x}}$, and show that $f(x)$ is maximum when $x=e$.

But here the domain of the set is $\mathbb{N}$. So how can I find the supremum of the above set?

Please anyone help me solve it. Thanks in advance.

Answer 1

You know that as a real function, $x^{1/x}$ is increasing on $(0, e)$ and decreasing on $(e, \infty)$. The same must be true if we consider it a function on the integers. That means that the maximum among integers must be at either $2$ or at $3$ (since any other integer input must give a function value strictly smaller than one of these two). Now just check those two input values, and you're done.

Answer 2

Try integers either side of $e$ - as the local maximum over the reals is in that area, the maximum over the natural numbers will be close by.

Answer 3

Note that $2^{1/2}=4^{1/4}$. Given what calculus tells you, you know the supremum is attained at $3$.

Answer 4

The first few values are $1$, $\sqrt 2=1.414\ldots$, $\sqrt[3]3=1.442\ldots$, $\sqrt2$, $\sqrt[5]5=1.379\ldots$, and we suspect that $\sqrt[3]3$ is the supremum (or in fact even maximum) of the sequence. It only remains to show that for larger $n$, we have $n^{1/n}<\sqrt[3]3$ or equivalently, $n^3<3^n$. This follows by induction after observing that the left hand side grows by a factor of $\frac{(n+1)^3}{n^3}=1+\frac 3n+\frac 3{n^3}+\frac1{n^3}\le 1+1+\frac13+\frac1{27}<3$ for $n\ge 3$.

Answer 5

Let $f(x)=x^{\frac{1}{x}}$ than $\ln f(x)=\frac{1}{x}\ln x$ and $\frac{f'(x)}{f(x)}=\frac{1}{x^2}-\frac{1}{x^2}\ln x$

$f'(x)=\frac{x^{\frac{1}{x}}}{x^2}(1-\ln x)<0$ for $x\geq 3$. This means our function is decreasing and we only need to check 3 values:

$1^1<\sqrt{2}<3^{\frac{1}{3}}$