Use the Maclaurin series 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 to evaluate ∑n=1∞(−1)n+1n⋅2n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \cdot 2^n}∑n=1∞n⋅2n(−1)n+1.
ln(2)\ln(2)ln(2)
ln(3/2)\ln(3/2)ln(3/2)
ln(2)2\frac{\ln(2)}{2}2ln(2)
ln(1/2)\ln(1/2)ln(1/2)