Answer
Yes, the given infinite geometric series has a limit, and the value of the limit is ${{S}_{\infty }}=\$25,000$.
Work Step by Step
The infinite geometric series is:
$\$500{{\left(1.02\right)}^{-1}}+\$500{{\left(1.02\right)}^{-2}}+\$500{{\left(1.02\right)}^{-3}}+\cdots$
Here ${{a}_{1}}=\$500{{\left(1.02\right)}^{-1}},{{a}_{2}}=\$500{{\left(1.02\right)}^{-2}}$
$\begin{align}
& \left| r \right|=\left| \frac{{{a}_{2}}}{{{a}_{1}}} \right| \\
& =\left| \frac{\$500{{\left(1.02\right)}^{-2}}}{\$500{{\left(1.02\right)}^{-1}}}\right|\\&=\left|{{\left(1.02\right)}^{-1}}\right|\\&={{\left(1.02\right)}^{-1}}\end{align}$
Find the limit of the infinite geometric series for the formula ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$.
$\begin{align}
& {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\
& =\frac{\$500{{\left(1.02\right)}^{-1}}}{1-{{\left(1.02\right)}^{-1}}}\\&=\frac{\frac{\$500}{\left(1.02\right)}}{1-\left(\frac{1}{1.02}\right)}\\&=\frac{\frac{\$500}{1.02}}{\frac{0.02}{1.02}}\end{align}$
This implies that,
$\begin{align}
& {{S}_{\infty }}=\frac{\$500}{1.02}\times\frac{1.02}{0.02}\\&=\frac{\$500}{0.02}\\&=\$25,000\end{align}$
Thus, the limit of the infinite geometric series is ${{S}_{\infty }}=\$25,000$.
