# Homework Help: Does the series converge?

1. Nov 15, 2009

### applegatecz

1. The problem statement, all variables and given/known data
Find all x for which $$\sum$$ from k=1 to infinity (x^k - x^(k-1))(x^k+x^(k-1)) converges.

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

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 15, 2009

### LCKurtz

Hint: Factor x(k-1) out.

3. Nov 16, 2009

### applegatecz

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

4. Nov 16, 2009

### 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.

5. Nov 16, 2009

### 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?

6. Nov 16, 2009

### LCKurtz

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?

7. Nov 16, 2009

### applegatecz

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

8. Nov 16, 2009

### 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?

9. Nov 16, 2009

### applegatecz

There is no fraction in the original summation.

10. Nov 16, 2009

### LCKurtz

Woops! Sorry, I misread the problem.