# Urgend Geometric series question

1. Feb 22, 2006

### Mathman23

Hi

I have the following problem:

show that

1/(1+x^2)) = 1-x^2 + x^4 + (-1)^n*(x^2n-2) + (-1)^n * (x^2n)/(1+x^2)

I that know this arctan function can be expanded as a geometric series by using:

1 + q + q^2 + q^3 + .... + = 1/(1-q)

Then by putting q = -x^2. I get:

1/(1-(-x^2) = 1 - x^2 - (-x^2) - (-x^2)^3 + .... +

My question is how do I proceed from this to get the desired result???

Sincerley and Best Regards

Fred

2. Feb 22, 2006

### arildno

No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2

3. Feb 22, 2006

### Mathman23

I still get first result, but has it something to do that the series converges? And then I then add a sum to the first result?

/Fred

4. Feb 22, 2006

### arildno

No, you simply haven't done it correctly. Try again.
Insert (-x^2) on the q-places in the following expression:
$$1+q+q^2+q^3+++$$
What do you get?

5. Feb 22, 2006

### assyrian_77

This is wrong. Try again.

6. Feb 22, 2006

### Mathman23

Hello by inserting q = -x^2 into the above I get:

1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... +

but how do I proceed from there ?

Sincerley

Fred

7. Feb 22, 2006

### assyrian_77

Now you got it right and should be able to "see" how the general term looks like. Just see how the sign changes and how the exponent increases.

8. Feb 22, 2006

### arildno

Great! One flaw though:
You've been working with 1/(1-(-x^2)), not 1/(1-x^2)!

Now, find the n'th term, and scrutinize the infinite part of the series starting with the n'th term.

9. Feb 22, 2006

### Mathman23

Hello I can see that the the function changes in the following way

(-1)^n * (x^2)^n, then

1/(1-x^2) = 1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... + (-1)^n * (x^2)^n

Do I then rewrite the above result to get the result required ???

Sincerely
Fred

10. Feb 22, 2006

### arildno

You have now found the general term of the series expansion of 1/(1-(-x^2))
For all N>=n, draw out the common factor (-1)^(n)x^(2n)
What are you left with then?

11. Feb 22, 2006

### Mathman23

The n-1 term ???

/Fred

12. Feb 22, 2006

### arildno

No.
Let's see:
We have:
$$(-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)$$
Try to deduce how the +++++ will look like..

13. Feb 22, 2006

### Mathman23

The n + 1 term ???

/Fred

14. Feb 22, 2006

### arildno

Try to figure out the rest by yourself

15. Feb 22, 2006

### Mathman23

I'v have been looking at mathworld, and then arrieved at the following idear:
Then I need to do it for infinity????

/Fred

$$S_n = \sum_{k=1} ^ {\infty} (-1)^k \frac{1}{1-x^k}$$

Last edited: Feb 22, 2006
16. Feb 22, 2006

### arildno

Try to tackle it this way:
The infinite series starting with the n'th term is of the form:
$$\sum_{i=n}^{\infty}(-1)^{i}x^{2i}$$
Now, introduce the new index j=i-n, that is i=j+n
Then, the series can be written in the form:
$$\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}$$

17. Feb 22, 2006

### Mathman23

I rewrite it as a product of a two Sums ??

/Fred

18. Feb 22, 2006

### arildno

No, draw out the common factor!
$$\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}$$
What is the j-sum equal to?

19. Feb 22, 2006

### Mathman23

??

$$\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1)^{n} x^{2n}$$

/Fred

Last edited: Feb 22, 2006
20. Feb 22, 2006

### arildno

Please write out the first few terms of the series $\sum_{j=0}^{\infty}(-1)^{j}x^{2j}$

Last edited: Feb 22, 2006