Consider the infinite geometric series ∑^∞n=1−4(1/3)^(n−1)

Question

In this image, the lower limit of the summation notation is "n = 1".

a. Write the first four terms of the series.
b. Does the series diverge or converge?
c. If the series has a sum, find the sum.

Answer 1

a. The first four terms of the series are:

n = 1: -4(1/3)^(1-1) = -4(1) = -4
n = 2: -4(1/3)^(2-1) = -4(1/3) = -4/3
n = 3: -4(1/3)^(3-1) = -4(1/9) = -4/9
n = 4: -4(1/3)^(4-1) = -4(1/27) = -4/27

b. To determine if the series converges or diverges, we need to calculate the common ratio of the geometric series. The common ratio is given by the formula r = -4(1/3)^(n-1) / -4(1/3)^(n-2)

r = -4(1/3)^1 / -4(1/3)^0
r = -4(1/3) / -4
r = 1/3

Since the absolute value of the common ratio is less than 1 (|1/3| < 1), the series converges.

c. To find the sum of the series, we use the formula for the sum of an infinite geometric series: S = a / (1 - r), where a is the first term and r is the common ratio.

a = -4
r = 1/3

S = -4 / (1 - 1/3)
S = -4 / (2/3)
S = -4 * (3/2)
S = -6

Therefore, the sum of the series is -6.