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

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

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

with radius of convergence R=1.

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

$\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n2^{n}}}=\ln(1+1/2)=\ln(3/2)$