Theorem: Limit of Error in Stirling's Formula

Consider Stirling's Formula:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
The ratio of $n!$ to its approximation $\sqrt {2 \pi n} \paren {\dfrac n e}^n$ is bounded as follows:
:$e^{1 / \paren {12 n + 1} } < \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } < e^{1 / 12 n}$

Proof:
Taking the natural logarithm of the ratio of $n!$ to its approximation $\sqrt {2 \pi n} \paren {\dfrac n e}^n$, we obtain:
{{begin-eqn}}
{{eqn | l = \map \ln {\dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } }
      | r = \map \ln {n!} - \map \ln { {\sqrt {2 \pi n} n^n e^{-n} } }
      | c = Difference of Logarithms
}}
{{eqn | r = \map \ln {n!} - \map \ln {\sqrt {2 \pi} } - \map \ln {\sqrt n} - \map \ln {n^n} - \map \ln {e^{-n} }
      | c = Sum of Logarithms
}}
{{eqn | r = \map \ln {n!} - \map \ln {\sqrt {2 \pi} } - \dfrac 1 2 \map \ln n- n \map \ln n + n
      | c = Logarithm of Power, Natural Logarithm of e is 1
}}
{{eqn | r = \paren {\ln n! - \paren {n + \dfrac 1 2} \ln n + n} - \map \ln {\sqrt {2 \pi} }
      | c = rearranging
}}
{{eqn | r = d_n - \map \ln {\sqrt {2 \pi} }
      | c = letting $d_n := \ln n! - \paren {n + \dfrac 1 2} \ln n + n$
}}
{{end-eqn}}
From the argument in Stirling's Formula: Proof 2: Lemma 3 we have that $\sequence {d_n - \dfrac 1 {12 n} }$ is an increasing sequence.
Then:
{{begin-eqn}}
{{eqn | l = d_n - d_{n + 1}
      | r = \ln n! - \paren {n + \frac 1 2} \ln n + n
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \paren {\ln \paren {n + 1}! - \paren {n + 1 + \frac 1 2} \map \ln {n + 1} + n + 1}
      | c = 
}}
{{eqn | r = -\map \ln {n + 1} - \paren {n + \frac 1 2} \ln n + \paren {n + \frac 3 2} \map \ln {n + 1} - 1
      | c = as $\ln \paren {n + 1}! = \map \ln {n + 1} + \ln n!$
}}
{{eqn | r = \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n} - 1
      | c = Difference of Logarithms
}}
{{eqn | r = \paren {\frac {2 n + 1} 2} \map \ln {\frac {2 n + 2} {2 n} } - 1
      | c = multiplying top and bottom by $2$
}}
{{eqn | r = \paren {\frac {2 n + 1} 2} \map \ln {\frac {\paren {2 n + 1} + 1} {\paren {2 n + 1} - 1} } - 1
      | c = adding and subtracting $1$
}}
{{eqn | r = \paren {\frac {2 n + 1} 2} \map \ln {\dfrac {\paren {2 n + 1} } {\paren {2 n + 1} } \times \paren {\frac {1 + \dfrac 1 {2 n + 1} } {1 - \dfrac 1 {2 n + 1} } } } - 1
      | c =
}}
{{eqn | n = 1
      | r = \paren {\frac {2 n + 1} 2} \map \ln {\frac {1 + \paren {2 n + 1}^{-1} } {1 - \paren {2 n + 1}^{-1} } } - 1
      | c = dividing top and bottom by $\paren {2 n + 1}$
}}
{{end-eqn}}
Let:
:$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$
for $\size x < 1$.
By Stirling's Formula: Proof 2: Lemma 1:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \sum_{n \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}
      | c = Lemma 1
}}
{{eqn | r = \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots
      | c = 
}}
{{eqn | o = >
      | r = \frac {x^2} 3 \sum_{n \mathop = 0}^\infty \paren {\dfrac {x^2} 3}^n
      | c = 
}}
{{eqn | n = 2
      | o = >
      | r = \frac {x^2} {3 \paren {1 - \dfrac {x^2} 3} } 
      | c = Sum of Infinite Geometric Sequence
}}
{{end-eqn}}
As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be substituted for $x$ in $(2)$:
{{begin-eqn}}
{{eqn | l = d_n - d_{n + 1}
      | o = \gt
      | r = \frac {\paren {\paren {2 n + 1}^{-1} }^2} {3 \paren {1 - \dfrac {\paren {\paren {2 n + 1}^{-1} }^2} 3} } 
      | c = 
}}
{{eqn | r = \frac {\paren {\paren {2 n + 1}^{-1} }^2} {3 \paren {1 - \dfrac {\paren {\paren {2 n + 1}^{-1} }^2} 3} } \times \frac {\paren {2 n + 1}^2} {\paren {2 n + 1}^2}
      | c = multiplying top and bottom by $\paren {2 n + 1}^2$
}}
{{eqn | r = \frac 1 {3 \paren {\paren {2 n + 1}^2 - \dfrac 1 3} }
      | c = simplifying
}}
{{eqn | r = \frac 1 {\paren {12 n^2 + 12 n + 2} }
      | c = 
}}
{{eqn | o = >
      | r = \frac 1 {\paren {12 n^2 + 14 n + \dfrac {13} {12} } }
      | c = key observation
}}
{{eqn | r = \frac 1 {12 \paren {n + \dfrac 1 {12} } \paren {n + \dfrac {13} {12} } }
      | c = 
}}
{{eqn | r = \frac 1 {12} \times \paren {\frac 1 {\paren {n + \dfrac 1 {12} } } - \frac 1 {\paren {n + \dfrac {13} {12} } } }
      | c = 
}}
{{eqn | r = \paren {\frac 1 {12 n + 1} - \frac 1 {12 n + 13} }
      | c = 
}}
{{eqn | r = \paren {\frac 1 {12 n + 1} - \frac 1 {12 \paren {n + 1} + 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = d_n - \frac 1 {12 n + 1} 
      | o = >
      | r = d_{n + 1} - \frac 1 {12 \paren {n + 1} + 1}
}}
{{end-eqn}}
Therefore the sequence $\sequence {d_n - \dfrac 1 {12 n + 1} }$ is decreasing.
From Stirling's Formula, we have that:
:$\ds \lim_{n \mathop \to \infty} d_n = \ln \sqrt {2 \pi}$
and so:
:$d_n - \dfrac 1 {12 n} < \ln \sqrt {2 \pi} < d_n - \dfrac 1 {12 n + 1}$
for $n = 1, 2, 3, \ldots$
Therefore:
{{begin-eqn}}
{{eqn | ll = - \dfrac 1 {12 n}
      | o = 
      | lo = <
      | l = -d_n + \ln \sqrt {2 \pi}
      | ro = <
      | rr = - \dfrac 1 {12 n + 1}
      | c = subtracting $d_n$ throughout
}}
{{eqn | ll = \dfrac 1 {12 n}
      | o = 
      | lo = >
      | l = d_n - \ln \sqrt {2 \pi}
      | ro = >
      | rr = \dfrac 1 {12 n + 1}
      | c = multiplying by $-1$
}}
{{eqn | ll = \dfrac 1 {12 n + 1}
      | o = 
      | lo = <
      | l = d_n - \ln \sqrt {2 \pi}
      | ro = <
      | rr = \dfrac 1 {12 n}
      | c = switching the order
}}
{{eqn | ll = e^{1 / \paren {12 n + 1} }
      | o = 
      | lo = <
      | l = \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} }
      | ro = <
      | rr = e^{1 / 12 n}
      | c = taking exponentials
}}
{{end-eqn}}
{{qed}}