Consider the infinite geometric series below.
a. Write the first 4 terms of the series
b. Does the series diverge or converge?
c. If the series has a sum, find the sum.
∞
∑ (-2)^n-1
n=2
To solve this problem, we'll examine the given infinite geometric series step by step.
a. To find the first four terms of the series, we substitute different values of n into the formula:
Starting with n = 2, the formula becomes: (-2)^(2-1) = (-2)^1 = -2
For n = 3: (-2)^(3-1) = (-2)^2 = 4
For n = 4: (-2)^(4-1) = (-2)^3 = -8
For n = 5: (-2)^(5-1) = (-2)^4 = 16
Therefore, the first four terms of the series are: -2, 4, -8, 16.
b. To determine if the series diverges or converges, we need to find the common ratio (r) of the geometric series. In this case, the common ratio is (-2), since each subsequent term is multiplied by this factor to obtain the next term.
For a geometric series to converge, the absolute value of the common ratio (|r|) must be less than 1. If |r| ≥ 1, the series diverges.
In this case, |r| = |-2| = 2, which is greater than 1. Therefore, the series diverges.
c. Since the series diverges, it does not have a finite sum.
∞
∑ (-2)^n-1
n=2
= -2 + 4 - 8 + 16 - ...
= (-2+4) + (-8+16) + (-32+64) ....
= 2 + 8 + 32 + ...
so what do you think ?