Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (a)/Solution 1

Recall that the Maclaurin series of $\ln (1+x)$ is

$\ln (1+x)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n}{x}^n$

with radius of convergence R=1.

If we put x=1/2, then we obtain the given series and hence we find

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n{2}^n}=\ln (1+1/2)=\ln (3/2)$