## Calculus 8th Edition

$1-1+1-1+\cdots+(-1)^{n-1}$
We expand the sigma notation by writing $(-1)^j$ as $j$ increases from 0 to n-1: $\displaystyle \sum_{j=0}^{n-1}(-1)^{j}$ $=(-1)^0+(-1)^1+(-1)^2+(-1)^3+(-1)^4+...+(-1)^{n-1}$ $=1-1+1-1+\cdots+(-1)^{n-1}$ (We see that most terms will cancel, but we don't know what the final answer will be without knowing whether n is even or odd.)