Question

determine whether the series (sum_{n = 0}^{infty}\frac{4^{n}-2^{n}}{5^{n}}) converges or diverges. if it converges, find its sum.
select the correct choice below and, if necessary, fill in the answer box within your choice.
a. the series converges because (lim_{n
ightarrowinfty}\frac{4^{n}-2^{n}}{5^{n}} = 0). the sum of the series is (square) (type an integer or a simplified fraction.)
b. the series diverges because (lim_{n
ightarrowinfty}\frac{4^{n}-2^{n}}{5^{n}}
eq0) or fails to exist
c. the series converges because it is the difference between two geometric series, each with (|r|lt1). the sum of the series is (square) (type an integer or a simplified fraction.)
d. the series diverges because it is the difference between two geometric series, at least one with (|r|geq1)

Explanation:

Step1: Split the series

Split \(\sum_{n = 0}^{\infty}\frac{4^{n}-2^{n}}{5^{n}}\) into \(\sum_{n = 0}^{\infty}\frac{4^{n}}{5^{n}}-\sum_{n = 0}^{\infty}\frac{2^{n}}{5^{n}}\), which is \(\sum_{n = 0}^{\infty}(\frac{4}{5})^{n}-\sum_{n = 0}^{\infty}(\frac{2}{5})^{n}\).

Step2: Use the geometric series sum formula

For a geometric series \(\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}\), when \(|r|\lt1\).
For \(\sum_{n = 0}^{\infty}(\frac{4}{5})^{n}\), \(r=\frac{4}{5}\), and its sum is \(\frac{1}{1-\frac{4}{5}}=\frac{1}{\frac{1}{5}} = 5\).
For \(\sum_{n = 0}^{\infty}(\frac{2}{5})^{n}\), \(r=\frac{2}{5}\), and its sum is \(\frac{1}{1-\frac{2}{5}}=\frac{1}{\frac{3}{5}}=\frac{5}{3}\).

Step3: Calculate the difference

The sum of \(\sum_{n = 0}^{\infty}(\frac{4}{5})^{n}-\sum_{n = 0}^{\infty}(\frac{2}{5})^{n}\) is \(5-\frac{5}{3}=\frac{15 - 5}{3}=\frac{10}{3}\).

Answer:

C. The series converges because it is the difference between two geometric series, each with \(|r|\lt1\). The sum of the series is \(\frac{10}{3}\)