Does the series converge for all values of x?

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Homework Statement


Find all x for which \sum from k=1 to infinity (x^k - x^(k-1))(x^k+x^(k-1)) converges.


Homework Equations


I think the geometric series formula is relevant: \sum k=N to infinity of x^k = 1/(1-x) for all |x|<1.


The Attempt at a Solution


I simplified the expression to x^2k - x^(2k-2). I can show that the first term converges (I think ... because it is the product of two convergent sequences?), and I understand logically why the second term converges, but not sure how to show rigorously. I also think that the series converges for all |x|<1, but again not sure how to construct the proof.
 
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Hint: Factor x(k-1) out.
 
But then x^2 is one of the factors, and x^2 does not converge (?).
 
You are summing over k, not over x.

Hint number 2: your guess for the values of x where the series converges is almost right, but not quite.
 
Ah, I see, thank you. In this case, unlike the "regular" geometric series case, x can be less than one OR one (i.e., |x|<=1), because if x is one, x^2k and x^(2k-1) are both one ... so the series converges to zero. Correct?
 
applegatecz said:
Ah, I see, thank you. In this case, unlike the "regular" geometric series case, x can be less than one OR one (i.e., |x|<=1), because if x is one, x^2k and x^(2k-1) are both one ... so the series converges to zero. Correct?

Yes, it converges at x = 1. But if you follow my hint and factor out x(k-1) in your original expression, there won't be any k left in the summand. What does that tell you?
 
OK, I think I understand: the expression factors to x^(k-1)*[x^(k-1)-x^(k-1)] = 0?
 
In the summation you started with

\frac {x^k - x^{k-1}}{x^k+x^{k-1}}

It's a fraction now, and it will be a fraction after you factor x(k-1) out and simplify it. Do that and simplify it. What do you get?
 
There is no fraction in the original summation.
 
  • #10
applegatecz said:
There is no fraction in the original summation.

Woops! Sorry, I misread the problem.
 
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