Help with Series: \Sum_{r=0}^{\infty}\frac{x^{2r}}{(1-x^{2r+1})(1-x^{2r+3})}

[maverick280857]

Hi

A friend gave me a problem the solution to which is unknown (I think somebody made it up and I am not sure if it is correct):

$$\frac{1}{(1-x)(1-x^3)} + \frac{x^2}{(1-x^3)(1-x^5)} + \frac{x^4}{(1-x^5)(1-x^7)} + ... (to \infty)$$

I figured that x must never equal one. Next, I can write this as

$$\sum_{r = 0}^{r = \infty} \frac{x^{2r}}{(1-x^{2r + 1})(1-x^{2r+3})}$$

I would like some help regarding

(a) the approach
(b) the correctness of the problem

I am unable to figure out how to proceed further. Tried partial fractions, but that didn't work. I am sort of stuck, so if you have any inputs to offer, I would be very grateful to listen to them...

Thanks and cheers
Vivek

EDIT: I understand that the range of x is necessary here: whether x is less than 1 or greater than 1. However, nothing has been mentioned in the problem.

 

[Nenad]

what is the problem. You have to proove that the first part is equal to the second part?

 

[matt grime]

What is the problem? to find the x for whichi t converges? to find a simple expresion for it, perhaps a closed form. it looks like it might be a generating function of some kind, there are somewhere tables to look these things up.

 

[maverick280857]

Thanks for your reply Matt and Nenad. I need hardly say that the first representation is indeed equivalent to the second one and that is nothing to prove. In fact the second method is a concise representation of the series in the form of the "r-th term". What I am looking for is a sum of this sequence for those values of x for which each term is defined.

I tried computing the series in Mathematica but all that the software does is report the input back. I am wondering why...

 

[matt grime]

Here's one method that might work or might just spew out something even more horrible.

let's forget convergence issues and treat it as a formal power series. so subs in the series for (1-x^r)^{-1} at all points and work out the coefficient of x^t. There are only a finite number of terms to add up (the r'th term of the sum only contributes powers of x^r and greater). perhaps these sum to something nice. perhaps not.

 

[maverick280857]

Here's one method that might work or might just spew out something even more horrible.

let's forget convergence issues and treat it as a formal power series. so subs in the series for (1-x^r)^{-1} at all points and work out the coefficient of x^t. There are only a finite number of terms to add up (the r'th term of the sum only contributes powers of x^r and greater). perhaps these sum to something nice. perhaps not.

I don't quite understand the approach you have proposed. Do you want me to compute the coefficient of $$x^t$$ in the entire series? What do you mean by "subs in the series for..."

Cheers
Vivek

 

[matt grime]

you know a series expansion for each term in the sum. put them all in and work out the general coefficient of a power of x., for any given power there are only a finite number of contributions to the sum.

the first term in the sum is (1+x+x^2+x^3+...)(1+x^3+x^6+x^9+...)
the second is x^2(1+x^3+x^6+x^9+...)(1+x^5+x^10+...)
and so on

work out the general coeff of a power of x and see if it's something nice, or if they satisfy some recurrence relations. it appears as though they might.

the whole sum starts off, after rearranging as

1+x+2^x^2+3x^3+...

 

[maverick280857]

Hi Matt

Yeah now I understand what you mean. I'll try and do this with a power series expansion.

But...there's got to be some neater way of doing this don't you think?

Thanks and cheers
Vivek

 

[shmoe]

Hmm, mathematica should be able to spit out a value for this if you give it a specific x. I'd doubt it's capable of actually finding a closed form though, but I could be wrong.

You can find the corresponding integral without too much trouble (or Mathematica should be able to do it for you). Maybe you can get some kind of asymptotic result, depending on the error terms when you switch to the integral.

 

[Zurtex]

I'm probably just stating the obvious, but if the pattern continues as:

$$x + 2x^2 + 3x^3 + \ldots + rx^r + \ldots$$

Then it is:

$$\frac{x}{(x-1)^2}$$

Assumedly for $|x|<1$

 

[uart]

The original series however converges for all real x, apart from x=1.

Looking at the behaviour of the tail you get two distinct cases, |x|<1 and |x|>1, both of which converge.

Case 1, |x|<1. In the tail the denominator "1" dominates the x^(r+1) and x^(r+3) so the denominator approaches 1 and the terms approach x^(2r), which converges (for |x|<1 ) by the root test.

Case 1, |x|>1. In the tail the denominator powers of x dominate the "1" and the terms approach x^(-2r-4) which again converges (for |x|>1) by the root test.

 

[robert Ihnot]

The original series however converges for all real x, apart from x=1.

Looking at the behaviour of the tail you get two distinct cases, |x|<1 and |x|>1, both of which converge.

Case 1, |x|<1. In the tail the denominator "1" dominates the x^(r+1) and x^(r+3) so the denominator approaches 1 and the terms approach x^(2r), which converges (for |x|<1 ) by the root test.

Case 1, |x|>1. In the tail the denominator powers of x dominate the "1" and the terms approach x^(-2r-4) which again converges (for |x|>1) by the root test.

If I could have got the latex to work, I would have written something more. At X=1, its infinity, and also at x=-1 since its an endless series of terms that are 1/4. So the problem is in general solved and you'd could evaluate it term by term for special values.

 

[maverick280857]

Well thanks for the inputs. I've had my share of trouble with problems requiring addition and subtraction of something just to get it to a standard form or to see a pattern. I was hoping I'd see something if I wrote it in the summation form. Didn't help :-D. Interestingly, I am told that one solution involves a similar "trick" so that some higher order terms tend to zero! I am trying to trace this problem right now and I'll post the actual solution once I get it, to continue the discussion.

Thanks and cheers,
vivek