consider the infinite geometric series infinity sigma n=1 -4(1/3)^n-1. In this, the lower limit of the summation notion 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.
Accepted Solution
A:
Answer:a) The first four terms are : -3 , -1/3 , 5/9 , 23/27b) The series convergesc) S∞ = (1)(n - 2) = ∞Step-by-step explanation:a) ∵ αn = 1 - 4(1/3)^n-1∴ If n = 1 ⇒ α1 = 1 - 4(1/3)^0 = -3∴ If n = 2 ⇒ α2 = 1 - 4(1/3)^1 = -1/3∴ If n = 3 ⇒ α3 = 1 - 4(1/3)^2 = 5/9∴ If n = 4 ⇒ α4 = 1 - 4(1/3)^3 = 23/27b) The series converges because with large number of n it approached to 1Note: If IrI < 1 then ⇒ convergec) S∞ = (1)(n - 2) ⇒ If n go to ∞ S∞ = ∞