Does the series converge for all values of x?

[applegatecz]

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.

 

[LCKurtz]

Hint: Factor x^(k-1) out.

 

[applegatecz]

But then x^2 is one of the factors, and x^2 does not converge (?).

 

[clamtrox]

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.

 

[applegatecz]

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?

 

[LCKurtz]

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?

 

[applegatecz]

OK, I think I understand: the expression factors to x^(k-1)*[x^(k-1)-x^(k-1)] = 0?

 

[LCKurtz]

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?

 

[applegatecz]

There is no fraction in the original summation.

 

[LCKurtz]

There is no fraction in the original summation.

Woops! Sorry, I misread the problem.