How to prove this inequality involving factorials?

Question

$$(n+1)\cdot ((n+1)!)^{\frac{1}{n+1}} -n\cdot (n!)^\frac{1}{n}< n+1$$ Where n is a positive integer. One post said that this can be done using Stirling's approximation, but how? Is there any other way this can be done?

Answer 1

Stirling approximation has a beauty within: it does work even for not so large $n$.

Stirling Approximation

Given $n!$, towards the Laplace methods for integrals, an approximation for the factorial reads:

$$n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$$

Where $e$ is the Euler Number.

Clearly this does extend to

$$(n+a)! \approx \sqrt{2\pi (n+a)} \left(\frac{n+a}{e}\right)^{n+a}$$

In our case $a = 1$.

The Inequality

First thing first, let's clean it a bit, dividing by $n+1$:

$$[(n+1)!]^{1/(n+1)} - \frac{n}{n+1}(n!)^{1/n} < 1$$

Now we apply Stirling approximation, and being careful about the exponents we are lead eventually to

$$[2 \pi (n+1)]^{1/(2(n+1))}\frac{n+1}{e} - \frac{n^2}{e(n+1)}(2\pi n)^{1/2n} < 1$$

Now some consideration according to what $n$ is, are mandatory.

Case 1: large $n$ values

In this hypothesis there is nothing wrong if we say $n+1 \approx n$, hence we get

$$(2\pi n)^{1/2n}\frac{n}{e} - \frac{n}{e}(2\pi n)^{1/2n} < 1$$

$$0 < 1$$

Which holds, and it clearly always holds for large values of $n$.

Case 2, small $n$ values

In this case we may take a series expansion, by considering small values of $n$.

This is more a curiosity factorial as it's meant, is only defined for natural numbers, hence "small" values of $n$ would be $n = 1$.

Probably by induction you may also be able to prove the inequality, after the Stirling approximation, but for the sake of maths, here is the small $n$ values.

By a series expansion, taking only the very first terms, we have

$$[2 \pi (n+1)]^{1/(2(n+1))} \approx \sqrt{2\pi} + \frac{n\sqrt{\pi}}{2}(1 - \ln(2\pi))$$

and

$$(2\pi n)^{1/2n} \approx (2\pi)^{1/2n}\ e^{\ln(n)/2n}$$

Substituting it into the inequality will clearly show that the second term is always greater than the first one, because of the exponential term, hence if you condense the writing as

$$A - B < 1$$

this will always hold.