Let

(1)

Multiplying by

(2)

Subtracting (2) from (1)



Factoring, and then dividing



Substituting into to form a new series,

One could obtain the same answer by initially defining the series as



and following a similar procedure.

I think there might be another solution maybe more visual.
Something like :



then multiplying on both sides :



and proceeding the multiplication :

.

As the series is endless all the unneeded parts are just gone, and there remains the fondamental equation :

chag-art:
This seems good to me. If you'd like to, you could post this as an alternative solution with the topic subject being, "Exercise [4.4] b" (without the inverted commas).

The equation/exercise 4.4 is flawed in the book. It valids only for |x| < 1.

The proof in here is flawed for the same reason.

The equation Sn - xSn = 1 is valid only for |x| < 1

For |x| >= 1, you'll get (Sn - xSn) -> (infinity - infinity*infinity) = infinity.

That is to say the author of this book should not have speculated with x = 2 then came up with the 'answer' -1/3 for the series. The Eq is not valid for x = 2 right from the start.

Well, Professor Penrose doesn't seem to discuss this issue in sufficient detail. He does assert that there are 'senses' in which the expression (where ) is 'correct'. He refers the reader to Hardy (1949) - this book discusses (I presume) many of the subtleties of divergent series.

In this exercise, we are asked to 'check' the expression given. The solution given here to Exercise [4.4] seems, to me, to be sufficient. In showing that the r.h.s. and l.h.s. are equivalent, in some sense, we would have checked the expression. There are definitely subtleties involved in how this equivalence is to be interpreted, but I think that these issues are beyond what is required in this exercise. I may be wrong!

This topic may be of interest:
http://www.physicsforums.com/showthread.php?t=221881

l.h.s and r.h.s are equivalent? What does 'equivalent' mean? 'A is equivalent to B' is written as A<=>B. If you mean 'equal', then lhs is definitely not equal to rhs, unless you supply a condition.

There may be something intersting about these sums, but the text here is definitely muddled. Anyway, I still don't see where -1/3 or -1/12 come from. In that wikipedia, your series is equal to -1/12(R), whatever R means.

The 'equivalence' here is derived from (what seems to be) sensible arguments. In my opinion, whatever is happening "behind the scenes" is beyond the scope of the requirements of the exercise. It would be useful to add these details though. I am not qualified to add details with regard to divergent series...yet. I urge anybody with the necessary background to elucidate this issue. It seems that there is more to this than merely stating:

• The equation Sn - xSn = 1 is valid only for |x| < 1

I think that the (R) signifies that the Ramanujan summation technique has been invoked.

This is more complex number magic!
If we restrict our domain to only real numbers then your argument is absolutely correct.But if we consider the domain to be the set of complex numbers then the entire scenario changes.A given function may converge even outside its circle of convergence!( although not every where)This is done formally by a procedure called ANALYTIC CONTINUATION.Check out this topic in chapter 7(section 7.4).
Adding one complex number to another does not give a 'bigger' or 'smaller' complex number.This is because the set of complex numbers is not 'ordered'.
The procedure of analytic continuation has mystic ties with the beautiful "Mandelbrot set" and chaos theory.

Sameed: Thank you for providing a flash of insight!

http://planetmath.org/encyclopedia/ExampleOfAnalyticContinuation.html

the correct answers to these sums are
(1-x^2^n)/(1-x^2), and (1-x^2^n)/(1+x^2), where n is number of members in the sum, and in this case is infinity. From that it is quite clear why they converge when IxI<1, and diverge when IxI>1. The plots for the solutions in the book are incorrect therefore, and I must say quite misleading. It is also quite clear from the above solutions why they are "similar" if x is a complex number.

can also be written down as a Taylor series equation. would this result in a proper solution for the exercise?

tulcod:

I had a similar idea when reading this exercise, but applying Mclaurin:

For doing so, one needs the successive derivatives of f(x) = 1/(1-x^2), for x=0. Here they are:



(Thanks to Vitalii Vanovschi, http://www.numberempire.com for his excellent derivatives calculator and other mathematical tools).

With the derivatives at hand it is straightforward to calculate the decomposition in series of the function:

1/(1-x^2) = 1 + x^2 + x^4 + x^6 + x^8 ….

I think this is an absolutely valid answer to the exercise. Checking is good but understanding is better.

Regards.

Manuel