Definition:Newton-Mercator Series

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

Then:

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 $-1 < x \le 1$.

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