# 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 $\map \ln {\paren {n + 1}!} = \map \ln {n + 1} + \map \ln {n!}$) $\ds$ $=$ $\ds \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n} - 1$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \frac {2n + 1} 2 \map \ln {\frac {1 + \paren {2 n + 1}^{-1} } {1 - \paren {2 n + 1}^{-1} } } - 1$

Let:

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

for $\size x < 1$.

Then:

 $\ds \map f x$ $=$ $\ds \sum_{k \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}$ Lemma 1 $\ds$ $<$ $\ds \frac {x^2} 3 \sum_{k \mathop = 0}^\infty x^{2 n}$ $\text {(1)}: \quad$ $\ds$ $<$ $\ds \frac {x^2} {3 \paren {1 - x^2} }$ Sum of Infinite Geometric Sequence

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

 $\ds d_n - d_{n - 1}$ $\le$ $\ds \frac 1 {3 \paren {\paren {2 n + 1}^2 - 1} }$ simplifying $\ds$ $=$ $\ds \frac 1 {12 n} - \frac 1 {12 \paren {n + 1} }$ $\ds \leadsto \ \$ $\ds d_n - \frac 1 {12 n}$ $\le$ $\ds d_{n-1} - \frac 1 {12 \paren {n + 1} }$

Thus the sequence:

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

is increasing.

$\blacksquare$