Stirling's Formula/Proof 2/Lemma 1

Lemma

Let $\map f x$ be the real function defined on the open interval $\openint {-1} 1$ as:

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

Then:

$\ds \map f x = \sum_{k \mathop = 1}^\infty \dfrac {x^{2 n} } {2 n + 1}$

Proof

\(\ds \map f x\)

\(:=\)

\(\ds \frac 1 {2 x} \map \ln {\frac {1 + x} {1 - x} } - 1\)

for $\size x < 1$

\(\ds \)

\(=\)

\(\ds \frac 1 {2 x} \paren {\map \ln {1 + x} - \map \ln {1 - x} } - 1\)

Difference of Logarithms

\(\ds \)

\(=\)

\(\ds \frac 1 {2 x} \paren {\sum_{k \mathop = 1}^\infty \paren {\paren {-1}^{k - 1} \frac {x^k} k} - \sum_{k \mathop = 1}^\infty \paren {-\frac {x^k} k} } - 1\)

Power Series Expansion for $\map \ln {1 + x}$

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 1}^\infty \frac {x^{2 n} } {2 n + 1}\)

simplifying

$\blacksquare$

Sources

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