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.
show all steps
a. The first four terms of the series can be found by plugging in n = 1, n = 2, n = 3, and n = 4 into the formula:
When n = 1:
a1 = -4(1/3)^0 = -4
When n = 2:
a2 = -4(1/3)^1 = -4/3
When n = 3:
a3 = -4(1/3)^2 = -4/9
When n = 4:
a4 = -4(1/3)^3 = -4/27
Therefore, the first four terms of the series are -4, -4/3, -4/9, -4/27.
b. To determine whether the series converges or diverges, we need to check the common ratio. The common ratio (r) in an infinite geometric series is found by dividing any term by the term before it. In this case, the common ratio is -1/3.
For an infinite geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, |-1/3| = 1/3 < 1, so the series converges.
c. To find the sum of the infinite geometric 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.
Plugging in the values a = -4 and r = -1/3:
S = -4 / (1 - (-1/3))
S = -4 / (4/3)
S = -4 * 3/4
S = -3
Therefore, the sum of the infinite geometric series is -3.