Multivariable Calculus, 7th Edition

$\Sigma^{\infty}_{n=1}ar^{n-1} = a+ar+ar^{2}$ If $|r| \lt 1$ the geometric series is convergent $\Sigma^{\infty}_{n=1}ar^{n-1} = \frac{a}{1-r}$ $|r| \lt 1$ If $|r| \geq 1$ the geometric series is divergent $\Sigma^{\infty}_{n=1} \frac{10^{n}}{(-9)^{n-1}}$ rewrite $n$-th term of the geometric series in the form $ar^{n-1}$ $=\Sigma^{\infty}_{n=1}10\frac{10^{n-1}}{(-9)^{n-1}}$ $=\Sigma^{\infty}_{n=1}10(\frac{10}{-9})^{n-1}$ $=\Sigma^{\infty}_{n=1}10(-\frac{10}{9})^{n-1}$ First term, $a=10$ and the common ratio is $r= -\frac{10}{9}$ $|r|=|-\frac{10}{9}|$ $=\frac{10}{9}$ $=1.1111$ $\gt 1$ Since $|r| \gt 1$, the geometric series is divergent.