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

In summary, the sequence 3/4^(2k) is a mathematical sequence where each term is obtained by raising 3/4 to the power of 2k. To show that the sequence is convergent, we can use the limit comparison test or show that it is a geometric series with a common ratio less than 1. The sum of the sequence can be calculated using the formula for the sum of a geometric series, and the sequence can be simplified to (3/16)^k. This sequence has various real-life applications, such as modeling exponential decay processes and financial calculations.
  • #1
939
111
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
  • #2
It is correct, if the summation starts with k=1.

ehild
 
  • Like
Likes 1 person
  • #3
ehild said:
It is correct, if the summation starts with k=1.

ehild

Thanks! Yes, it does.
 

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

What is the sequence 3/4^(2k)?

The sequence 3/4^(2k) is a mathematical sequence where each term is obtained by raising 3/4 to the power of 2k. This means that the first term is 3/4^0, the second term is 3/4^2, the third term is 3/4^4, and so on.

How do you show that the sequence is convergent?

In order to show that the sequence 3/4^(2k) is convergent, we need to prove that the terms of the sequence approach a finite limit as k approaches infinity. This can be done by using the limit comparison test or by showing that the sequence is a geometric series with a common ratio less than 1.

What is the sum of the sequence 3/4^(2k)?

The sum of the sequence 3/4^(2k) can be calculated by using the formula for the sum of a geometric series: S = a/(1-r), where a is the first term and r is the common ratio. In this case, a = 3/4^0 = 1 and r = 3/4^2 = 9/16. Therefore, the sum is S = 1/(1-9/16) = 16/7.

Can the sequence be written in a simplified form?

Yes, the sequence 3/4^(2k) can be written in a simplified form as (3/16)^k. This is because 3/4^2 = 3/16, and raising this value to the power of 2k is equivalent to multiplying 3/16 by itself k times.

What are some real-life applications of this sequence?

This sequence can be used to model exponential decay processes, such as radioactive decay or bacterial growth. It can also be applied in financial calculations, such as compound interest or amortization of loans.

Similar threads

Values of x for which a geometric series converges
Replies
1
Views
4K
How to show that these sequences are summable?
Replies
1
Views
911
What Values of x Allow Convergence in a Geometric Sequence of Sine Functions?
Replies
16
Views
2K
Finding Limit: Homework Statement & Solution
Replies
10
Views
2K
Using two sums to find geometric sequence
Replies
8
Views
2K
Prove that this series converges uniformly on R
Replies
4
Views
910
What is the lowest sum for K=2 in this infinite sequence with specific criteria?
Replies
2
Views
1K
General Formula for Finding Sum of Complex Sequence | Homework Help"
Replies
9
Views
2K
MHB No problem, happy to help! (Glad to hear it)
Replies
2
Views
1K
Show that the sequence converges
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top