Chapter 9 - Infinite Series - 9.2 Exercises - Page 601: 4

$1,\ \displaystyle \frac{3}{2},\ \displaystyle \frac{7}{4},\ \displaystyle \frac{23}{12},\ \displaystyle \frac{49}{24}$

$S_{1}=1$ $S_{2}=1+\displaystyle \frac{1}{2}=\frac{3}{2}$ $S_{3}=\displaystyle \frac{3}{2}+\frac{1}{4}=\frac{7}{4}$ $S_{4}=\displaystyle \frac{7}{4}+\frac{1}{6}=\frac{23}{12}$ $S_{5}=\displaystyle \frac{23}{12}+\frac{1}{8}=\frac{49}{24}$