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Calc 2 Sum of Alternating Geometric Series

  • Thread starter bigbob123
  • Start date
Problem 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
 

pasmith

Homework Helper
1,668
369
You can show that [itex](A_n)_{n \geq 0}[/itex] satisfies a second order linear recurrence of the form [tex]
A_{n+2} + pA_{n+1} + qA_n = 0,[/tex] which has general solution [tex]
A_n = C\lambda_1^n + D \lambda_2^n[/tex] where [itex]C[/itex] and [itex]D[/itex] are constants determined by the values of [itex]a_0[/itex] and [itex]a_1[/itex] and [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] are the roots of [tex]
\lambda^2 + p\lambda + q = 0.
[/tex]

You then have an expression for [itex]A_n[/itex] in closed form and can proceed to determine whether or not [itex]\sum_{n=0}^\infty A_n[/itex] converges.
 

Dick

Science Advisor
Homework Helper
26,252
615
If you look closely you'll notice that your series consists of two interlaced ordinary geometric series.
 
You can show that [itex](A_n)_{n \geq 0}[/itex] satisfies a second order linear recurrence of the form [tex]
A_{n+2} + pA_{n+1} + qA_n = 0,[/tex] which has general solution [tex]
A_n = C\lambda_1^n + D \lambda_2^n[/tex] where [itex]C[/itex] and [itex]D[/itex] are constants determined by the values of [itex]a_0[/itex] and [itex]a_1[/itex] and [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] are the roots of [tex]
\lambda^2 + p\lambda + q = 0.
[/tex]

You then have an expression for [itex]A_n[/itex] in closed form and can proceed to determine whether or not [itex]\sum_{n=0}^\infty A_n[/itex] converges.
It seems like the problem isn't linear though- An = 3An-2 / 4An-2
 
395
204
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.
 

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