Answer
Yes, the given infinite geometric series has a limit, and the value of the limit is ${{S}_{\infty }}=\frac{37}{99}$.
Work Step by Step
The infinite geometry series is:
$0.37+0.0037+0.000037+\cdots $
Here ${{a}_{1}}=0.37,{{a}_{2}}=0.0037$.
The value of $\left| r \right|$ is:
$\begin{align}
& \left| r \right|=\left| \frac{{{a}_{2}}}{{{a}_{1}}} \right| \\
& =\left| \frac{0.0037}{0.37} \right| \\
& =\left| 0.01 \right| \\
& =0.01
\end{align}$
Find the limit of the infinite geometric series for the formula ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$.
$\begin{align}
& {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\
& =\frac{0.37}{1-0.01} \\
& =\frac{0.37}{0.99} \\
& =\frac{37}{99}
\end{align}$
Thus, the limit of the infinite geometric series is ${{S}_{\infty }}=\frac{37}{99}$.
