Answer
a) $S_n=\dfrac{a(1-r^n)}{1-r}$
b) geometric; converges; $S=\dfrac{a}{1-r}$; for $|r|\geq 1$ the series diverges
Work Step by Step
a) The $n$th partial sum of a geometric sequence $a_n=ar^{n-1}$ is given by
$$S_n=\dfrac{a(1-r^n)}{1-r}.$$
b) The series $\sum_{k=1}^{\infty}ar^{k-1}$ is an infinite **geometric** series. If $|r|<1$, then the series converges and its sum is $\dfrac{a}{1-r}$. If $|r|\geq 1$, the series diverges.
