First-order homogeneous recurrence relation with variable coefficient

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/Recurrence_relation#Solving_first-order_non-homogeneous_recurrence_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?

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Ray Vickson
Homework Helper
Dearly Missed

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/Recurrence_relation#Solving_first-order_non-homogeneous_recurrence_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
Staff Emeritus
Homework Helper
Gold Member
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 $\displaystyle \ A_{n} = A_{0} \,?$
$$\begin{array}{rcl} v_1 &=& \frac{2}{1+q} v_0 \\ v_2&=& \frac{2}{1+q^2} v_1 = \frac{2^2}{(1+q)(1+q^2)} v_0 \\ v_3 &=& \frac{2}{1+q^3} v_2 = \frac{2^3}{(1+q)(1+q^2)(1+q^3)} v_0 \\ \vdots & \vdots &\hspace{4em} \vdots \end{array}$$ 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.