Can We Apply AM-GM?

Relevant wiki: Stirling's Formula

Instead of the usual notation for arithmetic progressions, I will let the initial term be $\displaystyle a+d$ for sake of elegance in following expressions.

It is trivial to show that, for an arithmetic progression with initial term $\displaystyle a+d$ and common difference $\displaystyle d$ , $\displaystyle A_n = \frac{2a+ (n+1) d}{2}$

The product of all the terms can have $\displaystyle d$ factored out of each individual term to obtain:

$\displaystyle d^n (\frac{a}{d}+1)(\frac{a}{d}+2)\cdots(\frac{a}{d} +n)$

By an abuse of notation, I will use the factorial in place of the gamma function for arbitrary real argument for simplicity.

$\displaystyle (G_n)^n = d^n \frac{(n+\frac{a}{d}+1)!}{(\frac{a}{d})!} \approx d^n \frac{\sqrt{2\pi ( n+\frac{a}{d}+1)}}{\sqrt{2 \pi \frac{a}{d}}} \times \frac{\left( \frac{n}{e}+\frac{a}{d e}+\frac{1}{e}\right)^{(n+\frac{a}{d}+1)}}{ \left( \frac{a}{d e} \right)^{\frac{a}{d}}}$

After taking the $\displaystyle n^{\text{th}}$ root of both sides, as $\displaystyle n \to \infty$ , all of the terms tend to unity except the ones which have exponent part that is independent of $\displaystyle n$ . So the limiting behaviour of $\displaystyle G_n$ is $\displaystyle d \times \frac{\frac{a}{d}+n+1}{e} = \frac{a + (n + 1)d}{e}$

Taking the quotient of the expressions yields $\displaystyle \frac{e}{2} \times \left( 1+ \frac{a}{a+(n+1)d} \right)$

Which obviously tends to $\displaystyle \frac{e}{2}$ as $\displaystyle n \to \infty$

Stirling's formula or Stolz Theorem is a nice approach .

I Did by Riemann's Sums.

On writing the limit take log both sides and use the standard riemann sums technique to evaluate