Stirling's Formula/Proof 2/Lemma 1

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Lemma

Let $f \left({x}\right)$ be the real function defined on the open interval $\left({-1 \,.\,.\, 1}\right)$ as:

$f \left({x}\right) := \dfrac 1 {2 x} \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1$


Then:

$\displaystyle f \left({x}\right) = \sum_{k \mathop = 1}^\infty \dfrac {x^{2 n} } {2n + 1}$


Proof

\(\displaystyle f \left({x}\right)\) \(:=\) \(\displaystyle \frac 1 {2 x} \ln \left({\frac {1 + x} {1 - x} }\right) - 1\) for $\left\vert{x}\right\vert < 1$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 x} \left({\ln \left({1 + x}\right) - \ln \left({1 - x}\right)}\right) - 1\) Difference of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 x} \left({\sum_{k \mathop = 1}^\infty \left({\left({-1}\right)^{k-1} \frac {x^k} k}\right) - \sum_{k \mathop = 1}^\infty \left({-\frac {x^k} k}\right)}\right) - 1\) Power Series Expansion for $\ln \left({1 + x}\right)$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}\) simplifying

$\blacksquare$


Sources