Use ln(1+x)=∑n=1∞(−1)n−1xnn\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n}ln(1+x)=∑n=1∞n(−1)n−1xn for ∣x∣<1|x|<1∣x∣<1.
