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.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.2$