## University Calculus: Early Transcendentals (3rd Edition)

Here, $r=-2, a=1$ The Nth partial sum of a geometric series can be found as: $s_n=\dfrac{a(1-r^n)}{1-r}$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ This is a geometric series with$|r| \gt 1$, so it diverges.