# Newton-Mercator Series

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 It has been suggested that this page or section be merged into Power Series Expansion for Logarithm of 1 + x. (Discuss)

## Theorem

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

Then:

 $\displaystyle \ln \left({1 + x}\right)$ $=$ $\displaystyle x - \dfrac {x^2} 2 + \dfrac {x^3} 3 - \dfrac {x^4} 4 + \cdots$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } n x^n$ $\quad$ $\quad$

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.