Answer
Yes, the given infinite geometric series has a limit, and the value of the limit is ${{S}_{\infty }}=\frac{49}{4}$.
Work Step by Step
$7+3+\frac{9}{7}+\cdots $
Here, ${{a}_{1}}=7$, ${{a}_{2}}=3$
The value of $\left| r \right|$ is,
$r=\frac{3}{7}$
Thus, the series does have a limit.
Find the limit of the infinite geometric series:
${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$.
$\begin{align}
& {{S}_{\infty }}=\frac{7}{1-\frac{3}{7}} \\
& =\frac{7}{\frac{4}{7}} \\
& =\frac{7\cdot 7}{4} \\
& =\frac{49}{4}
\end{align}$
Therefore, ${{S}_{\infty }}=\frac{49}{4}$.
