showing monotonicity of $\frac{n^{1/n}}{n}$ with only elementary tools

Question

I would like to show , that the sequence is monotonically decreasing.
I do not want to use any advanced math like Stirling formula nor derivatives.
I was wondering whether it is possible to show it by computing $a_{n+1}-a_{n}\ \ \text{or}\ \ \frac{a_{n+1}}{a_n}$.
I had no success so far in doing so.
Thanks for help

Answer 1

Consider the log of the terms:

$$a_n=\ln(\sqrt[n]n/n)=-\frac{n-1}n\ln(n)$$

By taking the ratio of these terms:

$$\frac{a_{n+1}}{a_n}=\frac{n^2}{n^2-1}\frac{\ln(n+1)}{\ln(n)}$$

It is clear this is greater than one, and hence $a_n$ is negative and increasing in magnitude, and hence monotone. Hence $\sqrt[n]n/n$ is monotone decreasing.

Answer 2

$$\frac {n^{1/n}}{n}>\frac {(n+1)^{1/(n+1)}}{n+1}\iff$$ $$\iff \left(\frac {n^{1/n}}{n}\right)^{n(n+1)}>\left(\frac {(n+1)^{1/(n+1)}}{n+1}\right)^{n(n+1)}\iff$$ $$\iff n^{1-n^2}>(n+1)^{-n^2}\iff$$ $$\iff n^{n^2-1}<(n+1)^{n^2}.$$

Answer 3

$n$ is increasing.

So $\frac{1}{n}$ is decreasing.

So $1-\frac{1}{n}$ is increasing.

Again, $n$ is increasing and larger than $1$, so $n^{1-\frac{1}{n}}$ is increasing. (The base is growing and so is the exponent, all while the base is at least $1$.)

So $\frac{1}{n^{1-\frac{1}{n}}}$ is decreasing. This is equal to $\frac{n^{1/n}}{n}$.