Answer
Interval of Convergence : $\ln 3 \lt x \lt \ln 5$ and $\Sigma_{n=1}^{\infty} (e^x-4)^n=\dfrac{1}{5-e^x}$
Work Step by Step
We notice that $\Sigma_{n=0}^{\infty} (e^x-4)^n= \Sigma_{n=0}^{\infty} (f(x))^n$
Remember that the series will converge when $|f(x||=|e^x-4| \lt 1$
This implies that $3 \lt e^x \lt 5$
and $ \ln 3 \lt x \lt \ln 5$
This means that the interval of convergence is: $\ln 3 \lt x \lt \ln 5$
Next, we will find the sum of the given geometric series.
$S=\dfrac{a}{1-r}=\dfrac{1}{5-e^x}$
$\implies \Sigma_{n=1}^{\infty} (e^x-4)^n=\dfrac{1}{5-e^x}$
