Stirling's Formula/Proof 2/Lemma 2

Theorem

The sequence $\sequence {d_n}$ defined as:

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

The proof strategy is to demonstrate that the sign of $d_n - d_{n + 1}$ is positive.

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

$=$

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

$\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 $

$=$

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

Let:

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

for $\size x < 1$.

Then from Lemma 1:

$(2): \quad \ds \map f x = \sum_{k \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}$

Thus $\map f x > 0$ for $\size x < 1$.

Putting $x = \dfrac 1 {2 n + 1}$ it can be seen that $(1)$ is $\map f {\dfrac 1 {2 n + 1} }$.

As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be seen that $(2)$ can be applied and so:

$\forall n \in \N: d_n - d_{n + 1} \ge 0$

Thus $\sequence {d_n}$ is a decreasing sequence.

$\blacksquare$