Sequence of 3/4^(2k). Show is convergent, find sum

AI Thread Summary
The sequence ak = 3/(4^(2k)) is shown to be convergent with a common ratio of r = 0.0625, which is less than 1. The sum of the series, starting from k=1, is calculated to be 0.2. The calculations for the first few terms confirm the convergence and the accuracy of the sum. The discussion concludes with confirmation that the summation notation is correctly applied.
939
Messages
110
Reaction score
2

Homework Statement



Consider the sequence ak = (3)/(4^(2k)). Show is convergent, find sum. Please check work.

Homework Equations



ak = 3/(4^2k)

let s {n} be the series associated with the sequence. Cannot write summation notation here, but k starts at 1 (k = 1 on bottom) and infinity on top.

The Attempt at a Solution



ak1 = 0.1875
ak2 = 0.01171875
ak3 = 0.000732421876
ak4 = 0.00004577636719

ak4/ak3 = 0.0625

Convergence: 1) r = 0.0625. r < 1, ∴ convergent
Sum: 2) (0.1875)/(1-0.0625) = 0.2. Sum is 0.2
 
Last edited:
Physics news on Phys.org
It is correct, if the summation starts with k=1.

ehild
 
  • Like
Likes 1 person
ehild said:
It is correct, if the summation starts with k=1.

ehild

Thanks! Yes, it does.
 
Back
Top