## Calculus: Early Transcendentals 8th Edition

$\displaystyle -1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}$
We expand the sigma notation by writing $\frac{2k-1}{2k+1}$ as $k$ increases from 0 to 4: $\displaystyle \sum_{k=0}^{4}\frac{2k-1}{2k+1}$ $\displaystyle =\frac{2*0-1}{2*0+1}+\frac{2*1-1}{2*1+1}+\frac{2*2-1}{2*2+1}+\frac{2*3-1}{2*3+1}+\frac{2*4-1}{2*4+1}$ $\displaystyle =-1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}$