Contents

# Contents

## Idea

The “Mercator series” (so named after its appearance in Mercator 1667) is the Taylor series of the natural logarithm around 1.

## Statement

###### Proposition

The Taylor series of the natural logarithm around $1 \in \mathbb{R}$ is the following series:

(1)\begin{aligned} \underoverset{n = 0}{\infty}{\sum} \tfrac{1}{n!} \left( \frac{d^n}{ d x^n} ln(1 + x) \right)_{\vert x = 0} x^n & \;\; = \;\; \underoverset{n = 1}{\infty}{\sum} \frac {(-1)^{n+1}} {n} x^n \\ & \;\; = \;\; x - \tfrac{1}{2} x^2 + \tfrac{1}{3} x^3 - \tfrac{1}{4} x^4 + \cdots \,. \end{aligned}

###### Proof

For the first two terms notice that

$ln(1 + x) \;\xrightarrow{x \to 0}\; ln(1) \,=\, 0$

and that the derivative of the natural logarithm is:

$\frac{d}{d x} \ln(1 + x) \;=\; \tfrac{1}{1+x} \;\xrightarrow{ x \to 0 }\; 1 \,.$

From here on, noticing for $k \in \mathbb{N}_+$ that:

$\frac{d}{d x} \left( \frac{1}{(1 + x)^k} \right) \;=\; - k \frac{1}{(1 + x)^{k+1}} \;\xrightarrow{x \to 0}\; - k$

we obtain for $n \in \mathbb{N}_+$, by induction:

\begin{aligned} \frac{d^n}{d x^n} ln(1 + x) & \;=\; \frac{d^{n-1}}{d x^{n-1}} \left( \frac{1}{1 + x} \right) \\ & \;=\; (n-1)! \cdot (-1)^{n-1} \frac{1}{(1 + x)^{n-1}} \\ & \;\xrightarrow{ x \to 0 }\; (n-1)! \cdot (-1)^{n+1} \end{aligned} \,.

Plugging this into the defining equation on the left of (1) and using

$\frac{(n-1)!}{n!} = \frac{1}{n}$

yields the claim.

## References

Apparently first published in:

but will have been known before that.