Question

Determine whether the series

[infinity]
Σ 3ⁿ⁻¹/4ⁿ converges or diverges. If it converges, find its sum
ⁿ⁼⁰
Select the correct choice below and, if necessary, fill in the answer box within your choice.
A. The series diverges because it is the difference between two geometric series, at least one with |r|≥1. The series converges because it is the difference between two geometric series, each with |r|<1. The sum of the
B. series is
C. The series diverges because limₙ₌[infinity] 3ⁿ⁻¹/4ⁿ fails to exist.
D. The series converges because limₙ₌[infinity] 3ⁿ⁻¹/4ⁿ = 0. The sum of the series is
E. The series diverges because limₙ₌[infinity] 3ⁿ⁻¹/4ⁿ ≠ 0

Answer 1

The series [infinity]Σ (3ⁿ⁻¹/4ⁿ) converges. Its sum is 1 which aligns with option B.

Step-by-step explanation:

The series [infinity]Σ 3ⁿ⁻¹/4ⁿ can be tested for convergence or divergence using the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. Let's apply the ratio test to the given series:

Using the ratio test, we have:
limₙ₌[infinity] |(3ⁿ⁻¹/4ⁿ⁺¹)/(3ⁿ⁻¹/4ⁿ)|
= limₙ₌[infinity] |3ⁿ⁻¹ * 4ⁿ/(4ⁿ⁻¹ * 3ⁿ⁺¹)|
= limₙ₌[infinity] |3ⁿ⁻¹ * 4ⁿ/4ⁿ⁻¹ * 3ⁿ⁺¹|
= limₙ₌[infinity] |(3ⁿ⁻¹/3ⁿ⁺¹) * (4ⁿ/4ⁿ⁻¹)|
= limₙ₌[infinity] |(3ⁿ⁻¹/3ⁿ⁺¹) * (4/4ⁿ⁻¹)|
= limₙ₌[infinity] |1/(3 * 3ⁿ⁺¹) * 4/(4ⁿ⁻¹)|
= limₙ₌[infinity] |4/(3 * 4)|
= 1/3.

Since the limit is less than 1, the series converges. To find the sum of the series, we can use the formula for the sum of a convergent geometric series:

Sum = a / (1 - r)

where a is the first term and r is the common ratio. In this case, a = 3⁰⁻¹/4⁰ = 1/4 and r = 3ⁿ⁻¹/4ⁿ = 3/4. Plugging in the values, we get:

Sum = (1/4) / (1 - 3/4) = (1/4) / (1/4) = 1.