Calc 2 Sum of Alternating Geometric Series

Click For Summary
SUMMARY

The discussion focuses on the calculation of the sum of an alternating geometric series defined by the recurrence relation A_n = 3A_{n-1} / 4A_{n-2}. The series satisfies a second-order linear recurrence of the form A_{n+2} + pA_{n+1} + qA_n = 0, leading to a general solution A_n = Cλ_1^n + Dλ_2^n, where C and D are constants based on initial values A0 and A1. The convergence of the series ∑A_n can be analyzed by recognizing that it consists of two interlaced ordinary geometric series.

PREREQUISITES
  • Understanding of second-order linear recurrence relations
  • Familiarity with geometric series and their convergence
  • Knowledge of characteristic equations and roots
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation of closed-form solutions for linear recurrence relations
  • Explore the convergence criteria for infinite series, particularly geometric series
  • Investigate the application of characteristic equations in solving recurrences
  • Practice problems involving interlaced series and their convergence properties
USEFUL FOR

Mathematics students, educators, and anyone interested in advanced series convergence, particularly those studying calculus or discrete mathematics.

bigbob123
Messages
2
Reaction score
0
Homework Statement
Suppose that An (from n = 0 to inf.) = {1/1, 3/1, 3/4, 9/4, 9/16, 27/16, 27/64, 81/64...} where we start with 1 and then alternate between multiplying by 3 and 1/4. Find the sum of An from n = 0 to n = inf.
Relevant Equations
Sn = A0(1-r)/(1-r) iff |r| < 1
A0 = 1
A1 = 3

3(An-1) / 4(An-2) = An
 
Physics news on Phys.org
You can show that (A_n)_{n \geq 0} satisfies a second order linear recurrence of the form <br /> A_{n+2} + pA_{n+1} + qA_n = 0, which has general solution <br /> A_n = C\lambda_1^n + D \lambda_2^n where C and D are constants determined by the values of a_0 and a_1 and \lambda_1 and \lambda_2 are the roots of <br /> \lambda^2 + p\lambda + q = 0.<br />

You then have an expression for A_n in closed form and can proceed to determine whether or not \sum_{n=0}^\infty A_n converges.
 
If you look closely you'll notice that your series consists of two interlaced ordinary geometric series.
 
  • Like
Likes   Reactions: WWGD
pasmith said:
You can show that (A_n)_{n \geq 0} satisfies a second order linear recurrence of the form <br /> A_{n+2} + pA_{n+1} + qA_n = 0, which has general solution <br /> A_n = C\lambda_1^n + D \lambda_2^n where C and D are constants determined by the values of a_0 and a_1 and \lambda_1 and \lambda_2 are the roots of <br /> \lambda^2 + p\lambda + q = 0.<br />

You then have an expression for A_n in closed form and can proceed to determine whether or not \sum_{n=0}^\infty A_n converges.
It seems like the problem isn't linear though- An = 3An-2 / 4An-2
 
Dick said:
If you look closely you'll notice that your series consists of two interlaced ordinary geometric series.

Which can be rearranged under certain conditions, which you probably ought to first prove.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
29
Views
6K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K