# Newton-Mercator Series

## Contents

## Theorem

Let $\ln x$ be the natural logarithm function.

Then:

\(\displaystyle \ln \left({1 + x}\right)\) | \(=\) | \(\displaystyle x - \dfrac {x^2} 2 + \dfrac {x^3} 3 - \dfrac {x^4} 4 + \cdots\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } n x^n\) |

The series converges to the natural logarithm (shifted by $1$) for $-1 < x \le 1$.

This is known as the **Newton-Mercator series**.

## Proof

From Sum of Infinite Geometric Progression, we know that:

- $\displaystyle \sum_{n \mathop = 0}^\infty x^n$ converges to $\dfrac 1 {1 - x}$

for $\left\vert{x}\right\vert <1$

which implies that:

- $\displaystyle \sum_{n \mathop = 0}^\infty (-1)^n x^n$ converges to $\dfrac 1 {1 + x}$

We also know from Definition:Natural Logarithm that:

- $\ln(x+1)=\displaystyle \int_0^x \frac {\mathrm dt} {1+t}$

Combining these facts, we get:

- $\ln(x+1)=\displaystyle \int_0^x \displaystyle \sum_{n \mathop = 0}^\infty (-1)^n t^n dt$

From Linear Combination of Integrals, we can rearrange this to

- $\displaystyle \sum_{n \mathop = 0}^\infty (-1)^n \displaystyle \int_0^x t^n dt$

Then, using Integral of Power:

- $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {(-1)^n} {n+1} x^{n+1} $

We can shift $n+1$ into $n$:

- $\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {(-1)^{n-1}} {n} x^{n} $

This is equivalent to:

- $ \displaystyle \sum_{n \mathop = 1}^\infty \dfrac {(-1)^{n+1}} {n} x^{n} $

Finally, we check the bounds $x=1$ and $x=-1$.

For $x=-1$, we get:

- $\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {(-1)^{n+1}} {n} (-1)^n$

$(-1)^{n+1}$ and $(-1)^n$ will always have different signs, which implies their product will be $-1$.

This means we get:

- $-\displaystyle \sum_{n \mathop = 1}^\infty \dfrac 1 n$

This is the harmonic series which we know to be divergent.

We then check $x=1$.

We get:

- $\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {(-1)^{n+1}} {n} $

This is the alternating harmonic series which we know to be convergent.

Therefore, we can conclude that:

- $\ln(x+1)=\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {(-1)^{n+1}} {n} x^{n}$ for $-1 < x \le 1$.

$\blacksquare$

## Examples

### Newton-Mercator Series: $\ln 2$

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

## Also known as

The **Newton-Mercator series** is also known as the **Mercator series**.

## Source of Name

This entry was named for Isaac Newton and Nicholas Mercator.

## Historical Note

The **Newton-Mercator Series** was discovered independently by both Isaac Newton and Nicholas Mercator.

However, it was also independently discovered by Grégoire de Saint-Vincent.