Newton-Mercator Series/Example/2

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Example of use of the Newton-Mercator Series

$\ln 2 = 1 - \dfrac 1 2 + \dfrac 1 3 - \dfrac 1 4 + \dfrac 1 5 - \cdots$


Proof

From the Newton-Mercator Series:

\(\displaystyle \map \ln {1 + x}\) \(=\) \(\displaystyle x - \dfrac {x^2} 2 + \dfrac {x^3} 3 - \dfrac {x^4} 4 + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n x^n\)

This is valid for $-1 < x \le 1$.

Setting $x = 1$:

\(\displaystyle \map \ln {1 + 1}\) \(=\) \(\displaystyle 1 - \dfrac {1^2} 2 + \dfrac {1^3} 3 - \dfrac {1^4} 4 + \cdots\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \ln 2\) \(=\) \(\displaystyle 1 - \dfrac 1 2 + \dfrac 1 3 - \dfrac 1 4 + \dfrac 1 5 - \cdots\)

$\blacksquare$


Sources