Using the Maclaurin series ln(1+u)=∑n=1∞(−1)n+1nun\ln(1+u) = \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} u^nln(1+u)=n=1∑∞n(−1)n+1un for ∣u∣<1|u| < 1∣u∣<1, find the sum ∑n=1∞(−1)n+1n⋅4n\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \cdot 4^n}n=1∑∞n⋅4n(−1)n+1.
14\frac{1}{4}41
ln(3)\ln(3)ln(3)
ln(54)\ln\left(\frac{5}{4}\right)ln(45)
ln(2)\ln(2)ln(2)