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 have $\Sigma_{n=0}^{\infty} (e^x-4)^n= \Sigma_{n=0}^{\infty} (f(x))^n$
The series will converge when $|f(x)|=|e^x-4| \lt 1$
So, $3 \lt e^x \lt 5 \implies \ln 3 \lt x \lt \ln 5$
Now, the sum of a given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{5-e^x}$
Thus, the Interval of Convergence is:
$\ln 3 \lt x \lt \ln 5$
and $\Sigma_{n=1}^{\infty} (e^x-4)^n=\dfrac{1}{5-e^x}$
