📂Calculus

Taylor Series of the Logarithm Function

Formula¹

The logarithm function’s Taylor series is as follows.

$$ \log (1+x) = \sum\limits_{n=1}^\infty \dfrac{(-1)^{n+1} x^{n}}{n}, \qquad |x| \lt 1 $$

Explanation

$x$ also holds for complex numbers. By a convenient change of variable we obtain:

$$ \log z = \sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n+1} (z-1)^{n}}{n}, \qquad |z - 1| \lt 1 $$

Derivation

When $|t| \lt 1$, $\dfrac{1}{1+t}$ is the limiting value of a geometric series.

$$ \dfrac{1}{1+t} = \sum\limits_{n=0}^{\infty} (-t)^{n}, \qquad |t| \lt 1 $$

The above series converges absolutely for $|t| \lt 1$. Therefore, under $|t| \lt 1$ we may interchange integration and the limit.

$$ \int_{0}^{x} \dfrac{1}{1+t}dt = \int_{0}^{x} \sum\limits_{n=0}^\infty (-t)^n dt = \sum\limits_{n=0}^\infty \int_{0}^{x} (-t)^n dt = \sum\limits_{n=0}^\infty \dfrac{(-1)^{n} x^{n+1}}{n+1} $$

On the other hand, the left-hand side is the logarithm by the differentiation rule for the logarithm function.

$$ \int_{0}^{x} \dfrac{1}{1+t}dt = \log (1+x) $$

Hence

$$ \log (1+x) = \sum\limits_{n=0}^\infty \dfrac{(-1)^{n} x^{n+1}}{n+1} = \sum\limits_{n=1}^\infty \dfrac{(-1)^{n+1} x^{n}}{n}, \qquad |x| \lt 1 $$

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