Answer
Yes, the given infinite geometric series has a limit, and the value of the limit is ${{S}_{\infty }}=\frac{43}{99}$.
Work Step by Step
The infinite geometric series is:
$0.43+0.0043+0.000043+\cdots $
Here ${{a}_{1}}=0.43,{{a}_{2}}=0.0043$
The value of $\left| r \right|$ is,
$\begin{align}
& \left| r \right|=\left| \frac{{{a}_{2}}}{{{a}_{1}}} \right| \\
& =\left| \frac{0.0043}{0.43} \right| \\
& =\left| 0.01 \right| \\
& =0.01
\end{align}$
Find the limit of the infinite geometric series for the formula ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$.
$\begin{align}
& {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\
& =\frac{0.43}{1-0.01} \\
& =\frac{0.43}{0.99} \\
& =\frac{43}{99}
\end{align}$
Therefore, the limit of the infinite geometric series is ${{S}_{\infty }}=\frac{43}{99}$.
