Answer
False.
Work Step by Step
Since $\Sigma^{\infty}_{n=0}ar^n=\frac{a}{1-r}$ , $0\lt|r|\lt1$
Therefore, $\Sigma^{\infty}_{n=0}ar^n=a+\Sigma^{\infty}_{n=1}ar^n=(\frac{a}{1-r})$
$\Rightarrow\Sigma^{\infty}_{n=1}ar^n=(\frac{a}{1-r})-a$
The formula requires that the geometric series begin
with $n = 0$.
