Calc 2 Sum of Alternating Geometric Series

[bigbob123]

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

 

[pasmith]

You can show that $(A_n)_{n \geq 0}$ satisfies a second order linear recurrence of the form $$A_{n+2} + pA_{n+1} + qA_n = 0,$$ which has general solution $$A_n = C\lambda_1^n + D \lambda_2^n$$ 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 $$\lambda^2 + p\lambda + q = 0.$$

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.

 

[Dick]

If you look closely you'll notice that your series consists of two interlaced ordinary geometric series.

 

[bigbob123]

You can show that $(A_n)_{n \geq 0}$ satisfies a second order linear recurrence of the form $$A_{n+2} + pA_{n+1} + qA_n = 0,$$ which has general solution $$A_n = C\lambda_1^n + D \lambda_2^n$$ 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 $$\lambda^2 + p\lambda + q = 0.$$

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

 

[RPinPA]

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.