TITLE: Proof that $\lim_{n\to\infty}\frac{e^{2n}(2n)!}{\sqrt{2n}(2n)^{2n}}$ converges without using Stirling's formula
QUESTION [1 upvotes]: Is it possible to prove that
$$L=\lim_{n\to\infty}\frac{e^{2n}(2n)!}{\sqrt{2n}(2n)^{2n}}$$ converges given that
$$T=\lim_{n\to\infty}\frac{2^{2n}n!^2}{\sqrt{n}(2n)!}$$
converges? But the restriction is that one must not use Stirling's formula.
My attempt:
$$2T=\lim_{n\to\infty}\frac{2^{2n+1}n!^2}{\sqrt{n}(2n)!}=\lim_{n\to\infty}\frac{e^{(2n+1)\ln 2}n!^2}{\sqrt{n}(2n)!}\lt \lim_{n\to\infty}\frac{e^{(2n+1)\ln 2}(2n)!}{\sqrt{n}(2n)!}=\lim_{n\to\infty}\frac{e^{(2n+1)\ln 2}}{\sqrt{n}}=\infty$$
Also, creating inequalities by changing $(2n)!$ to $(2n)^{2n}$ in the denominator results in the bound being zero, so that doesn't seem to help either...

REPLY [1 votes]: In "Algebra" vol. 2 by Chrystal,
published in 1900,
there is a derivation
of the "elementary" version of
Stirling's formula
on pages 368-372.
It is based on the expansion
$-\ln(1-z)
=\sum_{k=1}^{\infty} \dfrac{z^k}{k}
$
for $0 < z < 1$.
It shows,
on page 371, that
$$n!
\gt C e^{-n}n^{n+1/2}\exp\left(\dfrac1{12n}-\dfrac1{24n^2}\right)
$$
and
$$n!
\lt C e^{-n}n^{n+1/2}\exp\left(\dfrac1{12n}+\dfrac1{24n(n-1)}\right)
$$
where
$$C
=\exp\left(1-\dfrac12\sum_{m=2}^{\infty} \dfrac{(m-1)(\zeta(m)-1)}{m(m+1)}\right).
$$
It then shows
in the usual way
using Wallis' product
that
$C
=\sqrt{2\pi}$.
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