Answer:
Converges
Step-by-step explanation:
Given series:
[tex]\displaystyle \sum^{\infty}_{k=1} \dfrac{2^{k} + 3^{k}}{4^{k}}[/tex]
To determine whether the given series converges or diverges, we can split it into two separate series:
[tex]\displaystyle \sum^{\infty}_{k=1} \dfrac{2^{k}}{4^{k}} + \sum^{\infty}_{k=1}\dfrac{3^{k}}{4^{k}}[/tex]
Simplify, by applying the power of a quotient exponent rule:
[tex]\displaystyle \sum^{\infty}_{k=1} \left(\dfrac{2}{4}\right)^k + \sum^{\infty}_{k=1}\left(\dfrac{3}{4}\right)^k \\\\\\\\ \sum^{\infty}_{k=1} \left(\dfrac{1}{2}\right)^k + \sum^{\infty}_{k=1}\left(\dfrac{3}{4}\right)^k[/tex]
The first component series is a geometric series with first term a = 1/2 and common ratio r = 1/2. The series converges if ∣r∣ < 1. As |1/2| < 1, this series converges.
The second component series is a geometric series with first term a = 3/4 and common ratio r = 3/4. The series converges if ∣r∣ < 1. As |3/4| < 1, this series also converges.
Since both component series converge, their sum also converges. Therefore, the given series:
[tex]\Large\boxed{\boxed{\sf converges}}[/tex]
