upper bound of a function $n^{1/\log(n)}$

Question

I have the following expression
$n^{1/\log(n)}, \quad where \quad n \in [1, 10,000]$.
When I solve this numericall, I get the resultant value 2.718282 for all $n \in [2, 10,000]$. On this basis, I can consider the upper bound 3, means that
$n^{1/\log(n)} < 3 \quad \forall \, n \in [1, 10000]$. The question is that how I can analytically show that the upper bound of this expression is less than 3.

Answer 1

We use the well-known property that $$a^b=e^{b\ln a}$$ where $e$ is the mathematical constant equal to approximately 2.71. Applying to this context, $$n^{1/\ln n}=e^{(ln n)/(\ln n)}=e^1=e$$

Answer 2

We have $$f (x) = x^{1/\log x} = e^{\log x \cdot 1/\log x} \equiv e.$$ So, $f$ is a constant function and for all $x \in (0, +\infty) - \{1\}$ its value is $e = 2.71828\cdots$.

Answer 3

$$e=1+\frac11+\frac12+\frac1{3!}+\frac1{4!}+\frac1{5!}\cdots<1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\frac1{2^4}\cdots=3.$$