Can the Convergence of a Sequence be Proven by Using Epsilon and Inequalities?

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Hello Physics Forums community! I've been struggling for a while with this one
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; so basically a sequence a_n is given to us such that the sequence b_n defined by b_n = pa_n + qa_(n+1) is convergent where abs(p)<q. I need to prove a_n is convergent also. Any hint would be of so much help, thank you.

(I've tried proving it is Cauchy but no insight ever comes, just messing with inequalities. I haven't fully understood the role played by abs(p)<q)
 
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sebasvargasl said:
Hello Physics Forums community! I've been struggling for a while with this one
emoticon-0121-angry.gif
; so basically a sequence a_n is given to us such that the sequence b_n defined by b_n = pa_n + qa_(n+1) is convergent where abs(p)<q. I need to prove a_n is convergent also. Any hint would be of so much help, thank you.

(I've tried proving it is Cauchy but no insight ever comes, just messing with inequalities. I haven't fully understood the role played by abs(p)<q)

You can start by trying to think what could go wrong if ##|p|<q## isn't true. Suppose ##a_n=(-1)^n## and ##p=1## and ##q=1##.
 
I'm well aware of the counter examples when the inequality doesn't hold. I just can't find a way to prove the theorem; and to do this I need to fully understand the importance of the inequality. I can't see it intuitively
 
You can use the definition in terms of epsilon.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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