Stirling's Formula/Proof 2/Lemma 3

Lemma

Let $\sequence {d_n}$ be the sequence defined as:

$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$

Then the sequence:

$\sequence {d_n - \dfrac 1 {12 n} }$

is increasing.

Proof

We have:

$\ds d_n - d_{n + 1}$

$=$

$\ds \map \ln {n!} - \paren {n + \frac 1 2} \ln n + n$

$\ds $

$\, \ds - \, $

$\ds \paren {\map \ln {\paren {n + 1}!} - \paren {n + 1 + \frac 1 2} \map \ln {n + 1} + n + 1}$

$\ds $

$=$

$\ds -\map \ln {n + 1} - \paren {n + \frac 1 2} \ln n + \paren {n + \frac 3 2} \map \ln {n + 1} - 1$

(as $\map \ln {\paren {n + 1}!} = \map \ln {n + 1} + \map \ln {n!}$)

$\ds $

$=$

$\ds \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n } - 1$

Difference of Logarithms

$\ds $

$=$

$\ds \paren {\frac {2 n + 1} 2 } \map \ln {\frac {2 n + 2} {2 n} } - 1$

multiplying top and bottom by $2$

$\ds $

$=$

$\ds \paren {\frac {2 n + 1} 2 } \map \ln {\frac {\paren {2 n + 1} + 1} {\paren {2 n + 1} - 1} } - 1$

adding and subtracting $1$

$\ds $

$=$

$\ds \paren {\frac {2 n + 1} 2 } \map \ln {\dfrac {\paren {2 n + 1} } {\paren {2 n + 1} } \times \paren {\frac {1 + \dfrac 1 {2 n + 1} } {1 - \dfrac 1 {2 n + 1} } } } - 1$

$\text {(1)}: \quad$

$\ds $

$=$

$\ds \paren {\frac {2 n + 1} 2 } \map \ln {\frac {1 + \paren {2 n + 1}^{-1} } {1 - \paren {2 n + 1}^{-1} } } - 1$

dividing top and bottom by $\paren {2 n + 1}$

Let:

$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$

for $\size x < 1$.

Then:

$\ds \map f x$

$=$

$\ds \sum_{n \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}$

Lemma 1

$\ds $

$=$

$\ds \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots$

$\ds $

$<$

$\ds \frac {x^2} 3 \sum_{n \mathop = 0}^\infty x^{2 n}$

$\text {(2)}: \quad$

$\ds $

$<$

$\ds \frac {x^2} {3 \paren {1 - x^2} }$

Sum of Infinite Geometric Sequence

As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be substituted for $x$ in $(2)$:

$\ds d_n - d_{n + 1}$

$\lt$

$\ds \frac {\paren {\paren {2 n + 1}^{-1} }^2} {3 \paren {1 - \paren {\paren {2 n + 1}^{-1} }^2} }$

$\ds $

$=$

$\ds \frac {\paren {\paren {2 n + 1}^{-1} }^2} {3 \paren {1 - \paren {\paren {2 n + 1}^{-1} }^2} } \times \frac {\paren {2 n + 1}^2} {\paren {2 n + 1}^2}$

multiplying top and bottom by $\paren {2 n + 1}^2$

$\ds $

$=$

$\ds \frac 1 {3 \paren {\paren {2 n + 1}^2 - 1} }$

simplifying

$\ds $

$=$

$\ds \frac 1 {3 \paren {4 n^2 + 4 n} }$

$\ds $

$=$

$\ds \frac 1 {12 n \paren {n + 1} }$

$\ds $

$=$

$\ds \frac 1 {12 n \paren {n + 1} } \times \frac {12 } {12 }$

multiplying top and bottom by $12$

$\ds $

$=$

$\ds \frac {12} {12 n \times 12 \paren {n + 1} }$

$\ds $

$=$

$\ds \frac 1 {12 n} - \frac 1 {12 \paren {n + 1} }$

$\ds \leadsto \ \ $

$\ds d_n - \frac 1 {12 n}$

$\lt$

$\ds d_{n + 1} - \frac 1 {12 \paren {n + 1} }$

Thus the sequence:

$\sequence {d_n - \dfrac 1 {12 n} }$

is increasing.

$\blacksquare$

Sources

1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.2$
1955: Herbert Robbins: A Remark on Stirling's Formula (Amer. Math. Monthly Vol. 62: pp. 26 – 29) www.jstor.org/stable/2308012