Consider the infinite geometric series ∑∞n=1−4(13)n−1 . 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.
a. The first four terms of the series can be found by plugging in n = 1, 2, 3, and 4:
n = 1: -4(13)^0 = -4*1 = -4
n = 2: -4(13)^1 = -4*13 = -52
n = 3: -4(13)^2 = -4*169= -676
n = 4: -4(13)^3 = -4*2197 = -8788
b. To determine if the series converges or diverges, we need to calculate the common ratio r. The common ratio r is found by dividing any term by the term before it. In this case, r = -52 / -4 = 13.
For an infinite geometric series to converge, the absolute value of the common ratio must be less than 1. Since the absolute value of r (|13|) is greater than 1, the series diverges.
c. Since the series diverges, it does not have a sum.