Proof that $\lim_{n\to\infty}\frac{e^{2n}(2n)!}{\sqrt{2n}(2n)^{2n}}$ converges without using Stirling's formula 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...
 A: Since $LT\sqrt{2}=\lim_{n\to\infty}\dfrac{n!^2e^{2n}}{n^{2n+1}}$, $LT$ is finite iff the Stirling approximation is correct up to a multiplicative constant (i.e. we don't need to know the $\sqrt{2\pi}$ factor is correct), a hypothesis I'll call weak Stirling. Given any proof that if $T$ converges so does $L$, we have a proof of weak Stirling in the form of $LT$'s convergence. So let's try verifying $L\sqrt{2}/T$ converges, with a strategy that would also prove weak Stirling:$$\dfrac{L\sqrt{2}}{T}=\lim_{n\to\infty}f(n),\,f(n):=\left(\dfrac{(2n)!}{n!}\right)^2\left(\frac{e}{4n}\right)^{2n}.$$When $n$ increments from $k$ to $k+1$, $f$ multiplies by$$\left(\frac{k}{k+1}\right)^{2k}\left(\frac{e(2k+1)}{2k+2}\right)^2=\exp\left(1/(12k^2)-o(1/k^2)\right)$$(I've edited out a lot of tedious little-$O$ notation expanding logarithms), which implies the limit is finite (because $\sum_k1/k^2$ is finite). As noted already, this implies weak Stirling; indeed, a direct proof of weak Stirling can be done the same way.
Major edit: how did I get the above asymptotic form? The left-hand side's logarithm is$$\begin{align}-2k\ln(1+1/k)+2+2\ln(1-\tfrac{1}{2k+2})&=-2+\tfrac1k-\tfrac{2}{3k^2}+o(k^{-2})\\&+2-\tfrac{1}{k+1}-\tfrac{1}{(2k+2)^2}-o((k+1)^{-2}).\end{align}$$The $\pm2$s cancel, and $\tfrac1k-\tfrac{1}{k+1}=\tfrac{1}{k(k+1)}\sim\tfrac{1}{k^2}$. Since $o((k+1)^{-2})=o(k^{-2})$ and $\tfrac{1}{(2k+2)^2}\sim\tfrac{1}{4k^2}$, the right-hand simplifies to$$\tfrac{1}{k^2}(1-\tfrac23-\tfrac14)+o(k^{-2})\sim\tfrac{1}{12k^2}.$$(Of course, our convergence argument doesn't care about the exact coefficient, just that the constant & $1/k$ terms vanish.)
A: This formula is quite easy to prove $$\left(\frac n3\right)^n<n!<\left(\frac n2\right)^n$$ and generally suffice to prove many results involving factorial, yet here you'll get $0<L<\infty$ with it, not really helping. 
I don't see how you can get the precise $\sqrt{n}$ term without using Stirling or at least an equivalent $\sim c\sqrt{n}\left(\frac ne\right)^n$ with some undetermined constant $c$...
You can have that by setting $u_n=\dfrac{n!}{\sqrt{n}}\left(\dfrac ne\right)^n\quad $ and $\quad v_n=\ln(u_n)$
$v_{n+1}-v_n=\ln\left(\frac{u_{n+1}}{u_n}\right)=\ln\left(e\cdot\left(\frac{n}{n+1}\right)^{n+\frac 12}\right)=1-(n+\frac 12)\ln(1+\frac 1n)=\cdots=O\left(\frac 1{n^2}\right)$
Since $\sum\frac 1{n^2}$ converges, then $(v_n)_n$ converges too (by telescoping sum) and so does $u_n\to c$.
A: 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
$C
=\sqrt{2\pi}$.
