What is the Maclaurin series for ln(1+x)\ln(1+x)ln(1+x)?
∑n=1∞xnn=x+x22+x33+⋯\sum_{n=1}^{\infty} \frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots∑n=1∞nxn=x+2x2+3x3+⋯
∑n=1∞(−1)n+1nxn=x−x22+x33−⋯\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots∑n=1∞n(−1)n+1xn=x−2x2+3x3−⋯
∑n=0∞xnn!=1+x+x22!+⋯\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots∑n=0∞n!xn=1+x+2!x2+⋯
∑n=0∞(−1)n(2n+1)!x2n+1=x−x33!+x55!−⋯\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots∑n=0∞(2n+1)!(−1)nx2n+1=x−3!x3+5!x5−⋯