Use the series 11+t=∑n=0∞(−1)ntn\frac{1}{1+t} = \sum_{n=0}^{\infty} (-1)^n t^n1+t1=∑n=0∞(−1)ntn to find ∫0x11+tdt\int_0^x \frac{1}{1+t} dt∫0x1+t1dt.
∑n=1∞(−1)n−1nxn\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}x^n∑n=1∞n(−1)n−1xn
∑n=0∞(−1)nxn\sum_{n=0}^{\infty} (-1)^n x^n∑n=0∞(−1)nxn
∑n=1∞(−1)nnxn\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}x^n∑n=1∞n(−1)nxn
∑n=0∞(−1)nxnn!\sum_{n=0}^{\infty} (-1)^n \frac{x^{n}}{n!}∑n=0∞(−1)nn!xn