If f(x)=∑n=1∞xnnf(x) = \sum_{n=1}^{\infty} \frac{x^n}{n}f(x)=∑n=1∞nxn on the interval ∣x∣<1|x| < 1∣x∣<1, what is f′(x)f'(x)f′(x)?
∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}∑n=0∞xn=1−x1
∑n=2∞xn−1n\sum_{n=2}^{\infty} \frac{x^{n-1}}{n}∑n=2∞nxn−1
∑n=1∞xn−1=11−x\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}∑n=1∞xn−1=1−x1
∑n=1∞xn−1n(n−1)\sum_{n=1}^{\infty} \frac{x^{n-1}}{n(n-1)}∑n=1∞n(n−1)xn−1