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.
since we want an infinite series, the common ratio must be less than 1. Once again, it appears we have a user who cannot type fractions or exponents. The series must therefore be
∞
∑∞ −4(1/3)^(n−1) = -4 - 4/3 - 4/9 - 4/27 - ...
n=1
it converges because r = 1/3 < 1
The sum is -4/(1 - 1/3) = -6
The leading - sign may also be just an artifact of the bad typing
a. The first four terms of the series are: -4, -4/3, -4/9, -4/27.
b. The series converges because the common ratio, which is (1/3), is between -1 and 1.
c. The sum of the series can be calculated using 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, we get:
S = -4 / (1 - 1/3) = -4 / (2/3) = -6.
Therefore, the sum of the series is -6.