Answer
Convergent
Work Step by Step
Here, we have: $\Sigma_{k=1}^{\infty}\dfrac{2^k+3^k}{4^k}=\Sigma_{k=1}^{\infty} (\dfrac{1}{2})^k+\Sigma_{k=1}^{\infty} (\dfrac{3}{4})^k$
We will use the Geometric series formula $\Sigma_{n=0}^{\infty} ar^n=a+ar+ar^2+......=\dfrac{a}{1-r}$
Now, $\Sigma_{k=1}^{\infty}\dfrac{2^k+3^k}{4^k}=\Sigma_{k=1}^{\infty} (\dfrac{1}{2})^k+\Sigma_{k=1}^{\infty} (\dfrac{3}{4})^k$
or, $=\dfrac{1/2}{1-\dfrac{1}{2}}+\dfrac{3/4}{1-\dfrac{3}{4}}$
or, $=1+3$
or, $\Sigma_{k=1}^{\infty}\dfrac{2^k+3^k}{4^k}=4$
This implies that the series is convergent by using Geometric series formula.
