Let $a_n = \dfrac {n!} {\sqrt {2 n} \paren {\frac n e}^n}$. === Part 1 === It will be shown that: :$\lim_{n \mathop \to \infty} a_n = a$ for some constant $a$. This will imply that: :$\displaystyle \lim_{n \mathop \to \infty} \frac {n!} {a \sqrt{2 n} \paren {\frac n e}^n} = 1$ By applying Power Series Expansion for $\map \ln {1 + x}$: {{begin-eqn}} {{eqn | l = \map \ln {\frac {n + 1} n}       | r = \map \ln {1 + \frac 1 {2 n + 1} } - \map \ln {1 - \frac 1 {2 n + 1} }       | c =  }} {{eqn | r = 2 \sum_{k \mathop = 0}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k + 1}       | c =  }} {{end-eqn}} Let $b_n = \ln a_n$. {{begin-eqn}} {{eqn | l = b_n - b_{n + 1}       | r = \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n} - 1       | c =  }} {{eqn | n = 1       | r = \sum_{k \mathop = 1}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k}       | c =  }} {{eqn | o = >       | r = 0       | c =  }} {{eqn | l = b_n       | o = >       | r = b_{n + 1}       | c =  }} {{end-eqn}} so the sequence $\sequence {b_n}$ is (strictly) decreasing. From $(1)$: {{begin-eqn}} {{eqn | l = b_n - b_{n + 1}       | r = \sum_{k \mathop = 1}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k}       | c =  }} {{eqn | ll= \leadsto       | l = b_n - b_{n + 1}       | o = <       | r = \sum_{k \mathop = 1}^\infty \paren {\frac 1 {2 n + 1} }^{2 k}       | c = as $\dfrac 1 {2 k + 1} < 1$ }} {{eqn | r = \frac 1 {4 n \paren {n + 1} }       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = b_1 - b_n       | r = \sum_{m \mathop = 1}^{n - 1} b_m - b_{m + 1}       | c = {{Defof|Telescoping Series}} }} {{eqn | o = <       | r = \frac 1 4 \sum_{m \mathop = 1}^{n - 1} \frac 1 {m \paren {m + 1} }       | c =  }} {{eqn | o = <       | r = \frac 1 4 \sum_{m \mathop = 1}^\infty \frac 1 {m \paren {m + 1} }       | c =  }} {{eqn | r = \frac 1 4       | c = Sum of Sequence of Products of Consecutive Reciprocals }} {{end-eqn}} {{explain|The use of Telescoping Series is unclear}} and so:  :$b_n > b_1 - \dfrac 1 4 = \dfrac 3 4 - \dfrac {\ln 2} 2$ Thus by definition $\sequence {b_n}$ is bounded below.  By Monotone Convergence Theorem, it follows that $\sequence {b_n}$ is convergent. Let $b$ denote its limit.  Then: :$\displaystyle \lim_{n \mathop \to \infty} a_n = e^{\lim_{n \mathop \to \infty} b_n} = e^b = a$ as required. {{qed|lemma}} === Part 2 === By Wallis's Product, we have: :$\displaystyle \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1} = \frac \pi 2$ or equivalently: :$(2): \quad \displaystyle \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {2 n + 1} } = \frac \pi 2$ In Part 1 it was proved that: :$n! \sim a \sqrt {2 n} \paren {\dfrac n e}^n$ Substituting for $n!$ in $(2)$ yields: {{begin-eqn}} {{eqn | l = \frac \pi 2       | r = \lim_{n \mathop \to \infty} \frac {2^{4 n} a^4 \paren {4 n^2} n^{4 n} e^{-4 n} } {a^2 \paren {4 n} 2^{4 n} n^{4 n} e^{-4 n} \paren {2 n + 1} }       | c =  }} {{eqn | r = a^2 \lim_{n \mathop \to \infty} \frac n {2 n + 1}       | c = as most of it cancels out }} {{eqn | r = \frac {a^2} 2       | c =  }} {{eqn | ll= \leadsto       | l = a       | r = \sqrt \pi       | c =  }} {{end-eqn}} Hence the result. {{qed}}