Find Sum of Series Using Power Series

1. Jun 28, 2011

Rapier

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

Find the sum of sigma (n=1 to infinity) (-1)^(n-1) * [n/(4^(n-1))] using the power series 1/(1+x) = sigma (n=0 to infinity) (-1)^n * x^n.

2. Relevant equations
1/(1+x) = 1 - x + x^2 - x^3 + x^4 + ... + (-1)^n * x^n

3. The attempt at a solution

The problem suggested to differentiate both sides and then substitute 1/4 for x.

I am totally confused and don't even know where to begin.

2. Jun 28, 2011

hunt_mat

You know how to differentiate right:
$$\frac{d}{dx}\frac{1}{1+x}=-\frac{1}{(1+x)^{2}}$$
I am sure you know how to differentiate $x^{n}$, what do you get if you do this?

3. Jun 28, 2011

Rapier

d/dx (x^n) = n * x^(n-1)

4. Jun 28, 2011

hunt_mat

Right, so if you differentiate the series term by term (which you're allowed to do as long as you're within the radius of convergence), what do you get?

5. Jun 28, 2011

Rapier

I'm not sure I understand the question.

Which series?

Maybe I'm just not understanding the original question.

1/(1+x) can be rewritten as sum (n=0 to infinity) of (-1)^n * x^n.

I believe the question is asking me to take the sum (n=1 to infinity) of (-1)^(n-1) * n/4^(n-1) and come up with a small simple equation to determine its sum using the pattern of 1/1+x as a template. Yes?

I know that if I take x^n and differentiate it I get n/x^n-1, which looks exactly like my n/4^(n-1). Does that mean that the sum of my series is nothing more than 1 - 1/4 + 1/16 - 1/64? Do I need to make an adjustment to my n values because my sum starts @1 and the original starts @0?

6. Jun 28, 2011

hunt_mat

Differentiating the series:
$$\sum_{n=0}^{\infty}(-1)^{n}x^{n}$$
is
$$\sum_{n=1}^{\infty}(-1)^{n}nx^{n-1}$$
Now what is the above equal to? (hint, we have derived it before)

7. Jun 28, 2011

vela

Staff Emeritus
Do exactly what was suggested. What do you get if you differentiate the lefthand side? What do you get if you differentiate the righthand side? Then plug in x=1/4. What do you get? How is it related to the original question?

8. Jun 28, 2011

Rapier

Since I differentiated the right side, I have to differentiate the left side, neh?

So that will equal -1/(1+x)^2?

I'm trying not to feel really stupid, and I'm just so frustrated. I greatly appreciate all your help.

EDIT: When you differentiate do you change the starting value for n? Since I'm not differentiating n (-1)^n stays (-1)^n, shouldn't it?

EDIT EDIT: Sum = - 16/25?

Last edited: Jun 28, 2011
9. Jun 28, 2011

hunt_mat

Correct, now following your suggestion, set x=1/4, what will the series look like?

10. Jun 28, 2011

Rapier

This is my differentiation:
$$\sum_{n=1}^{\infty}(-1)^{n}nx^{n-1}$$

The series was listed as:
$$\sum_{n=1}^{\infty}(-1)^{n-1}nx^{n-1}$$

Does the exponent on the (-1) change when differentiated?

But to answer your question, the series should be:

1 -1/2 + 3/16 - 1/16 + ...

11. Jun 28, 2011

hunt_mat

The idea here us to note that $(-1)^{n-1}=(-1)^{n+1}$ and you have a minus coming from the LHS.

12. Jun 28, 2011

Rapier

OH! I see that! If I divide (-1)^n by -1, I get (-1)^n-1. DUH!

That is how it comes up with being 1/(1+x)^2.

That makes sense.

13. Jun 28, 2011

hunt_mat

So, now you can tell us what the sum of the series is?

14. Jun 28, 2011

Rapier

16/25!!

15. Jun 28, 2011

hunt_mat

Well done!!!

16. Jun 28, 2011

Rapier

Thank you SO much for your patience and help. Not only did I understand what I did I can do it on other problems!