# Power Series Expansion for Logarithm of 1 + x

## Theorem

The Newton-Mercator series defines the natural logarithm function as a power series expansion:

 $\ds \map \ln {1 + x}$ $=$ $\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$ $\ds$ $=$ $\ds x - \frac {x^2} 2 + \frac {x^3} 3 - \frac {x^4} 4 + \cdots$

valid for all $x \in \R$ such that $-1 < x \le 1$.

### Corollary

 $\ds \map \ln {1 - x}$ $=$ $\ds -\sum_{n \mathop = 1}^\infty \frac {x^n} n$ $\ds$ $=$ $\ds -x - \frac {x^2} 2 - \frac {x^3} 3 - \frac {x^4} 4 - \cdots$

valid for $-1 < x < 1$.

## Proof

From Sum of Infinite Geometric Sequence, putting $-x$ for $x$:

$(1): \quad \ds \sum_{n \mathop = 0}^\infty \paren {-x}^n = \frac 1 {1 + x}$

for $-1 < x < 1$.

From Power Series Converges Uniformly within Radius of Convergence, $(1)$ is uniformly convergent on every closed interval within the interval $\openint {-1} 1$.

From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

 $\ds \int_0^x \frac 1 {1 + t} \rd t$ $=$ $\ds \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t}^n \rd t$ $\ds \leadsto \ \$ $\ds \map \ln {1 + x}$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{n + 1} } {n + 1}$ Primitive of Reciprocal and Integral of Power $\ds \leadsto \ \$ $\ds \map \ln {1 + x}$ $=$ $\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$ letting $n \to n - 1$

$\blacksquare$