First-order homogeneous recurrence relation with variable coefficient

[Robin04]

Homework Statement

I need to find the explicit formula for the following recursive sequence:
$v_n=\frac{2}{1+q^n}v_{n-1}$ where $0<q<1$ is a constant

Homework Equations

I found the following method to solve it:
https://en.wikipedia.org/wiki/Recur...currence_relations_with_variable_coefficients

The Attempt at a Solution

Referring to Wikipedia, first I need to find

, where

My sequence is homogenous so $g_n = 0$

I'm already in a problem at the beginning. In order to find $A_n$ I need to calculate$\prod_{k=0}^{n-1} \frac{2}{1+q^k} = 2^{n-1}\prod_{k=0}^{n-1} \frac{1}{1+q^k}$
I don't really have any idea how to do this. Can you give me some hint?

 

[Ray Vickson]

Homework Statement

I need to find the explicit formula for the following recursive sequence:
$v_n=\frac{2}{1+q^n}v_{n-1}$ where $0<q<1$ is a constant

Homework Equations

I found the following method to solve it:
https://en.wikipedia.org/wiki/Recur...currence_relations_with_variable_coefficients

The Attempt at a Solution

Referring to Wikipedia, first I need to find

, where

My sequence is homogenous so $g_n = 0$

I'm already in a problem at the beginning. In order to find $A_n$ I need to calculate$\prod_{k=0}^{n-1} \frac{2}{1+q^k} = 2^{n-1}\prod_{k=0}^{n-1} \frac{1}{1+q^k}$
I don't really have any idea how to do this. Can you give me some hint?

You are done; that is about as simple a formula as can possibly be found.

 

[SammyS]

You are done; that is about as simple a formula as can possibly be found.

Don't you also get the result that $\displaystyle \ A_{n} = A_{n-1} \ \dots \$ so that $\displaystyle \ A_{n} = A_{0} \,?$

 

[Ray Vickson]

Don't you also get the result that $\displaystyle \ A_{n} = A_{n-1} \ \dots \$ so that $\displaystyle \ A_{n} = A_{0} \,?$

Using the recursion directly (and not bothering to look up formulas in Wikipedia) we have
$$ \begin{array}{rcl}
v_1 &=& \frac{2}{1+q} v_0 \\
v_2&=& \frac{2}{1+q^2} v_1 = \frac{2^2}{(1+q)(1+q^2)} v_0 \\
v_3 &=& \frac{2}{1+q^3} v_2 = \frac{2^3}{(1+q)(1+q^2)(1+q^3)} v_0 \\
\vdots & \vdots &\hspace{4em} \vdots
\end{array} $$ So
$$v_n = 2^n \left( \prod_{k=1}^n \frac{1}{1+q^k} \right) \: v_0,$$ and it does not get simpler than that.