Answer
Yes, the given infinite geometric series has a limit, and the value of the limit is ${{S}_{\infty }}=\$12,500$.
Work Step by Step
$\$1000{{\left(1.08\right)}^{-1}}+\$1000{{\left(1.08\right)}^{-2}}+\$1000{{\left(1.08\right)}^{-3}}+\cdots$
Here ${{a}_{1}}=\$1000{{\left(1.08\right)}^{-1}},{{a}_{2}}=\$1000{{\left(1.08\right)}^{-2}}$
$\begin{align}
& \left| r \right|=\left| \frac{{{a}_{2}}}{{{a}_{1}}} \right| \\
& =\left| \frac{\$1000{{\left(1.08\right)}^{-2}}}{\$1000{{\left(1.08\right)}^{-1}}}\right|\\&=\left|{{\left(1.08\right)}^{-1}}\right|\\&={{\left(1.08\right)}^{-1}}\end{align}$
Find the limit of the infinite geometric series using the formula ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$.
$\begin{align}
& {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\
& =\frac{\$1000{{\left(1.08\right)}^{-1}}}{1-{{\left(1.08\right)}^{-1}}}\\&=\frac{\frac{\$1000}{\left(1.08\right)}}{1-\left(\frac{1}{1.08}\right)}\\&=\frac{\frac{\$1000}{1.08}}{\frac{0.08}{1.08}}\end{align}$
This implies that,
$\begin{align}
& {{S}_{\infty }}=\frac{\$1000}{1.08}\cdot\frac{1.08}{0.08}\\&=\frac{\$1000}{0.08}\\&=\$12,500\end{align}$
Thus, the limit of the infinite geometric series is ${{S}_{\infty }}=\$12,500$.
