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Ln2 = 0: series paradox?
By Andrew Smith (P2517) on Tuesday, May 23, 2000 - 08:30 pm:

This has been puzzling me for a while:
Clearly a problem.
When I first thought about this I thought the problem was with taking S from S because S is divergent so in effect S - S = ¥ - ¥ which is undefined but surely A is also divergent so S - 2A should not be defined but is (equal to ln2).

Can anyone give me an answer because anyone I've shown this to has used a lot of hand waving and not really given me an answer.

By Dan Goodman (Dfmg2) on Tuesday, May 23, 2000 - 09:15 pm:

The problem is that you can have two divergent series which added together termwise are convergent, but still make no sense. For instance, let An=S (-1)n, Bn=S (-1)n+1, so that An=(1,0,1,0,1,0,...) and Bn=(-1,0,-1,0,...). Adding termwise, you get Cn=An+Bn=0 which is convergent, so you might be tempted to say that A+B=0. However, it should also be the case that if A+B is defined then Dn=An+B2n+B2n+1 should also converge to A+B. But Dn=(0,-1,-2,-3,-4,...) which clearly diverges. So A+B is not defined even though it seems to have a value when adding it together termwise. The precise rule for when you can add two series together termwise is if both series converge absolutely. If the terms of the series are xn and yn, then you can add them together termwise if both S |xn| and S |yn| converge, i.e. that the sum of the positive values converges. So, with ln(2) you can't do this, because S |xn|=1+1/2+1/3+1/4+... which diverges.

By Simon Munday (Sjm78) on Tuesday, May 23, 2000 - 09:31 pm:

Andrew,

I'm afraid what you've written doesn't really mean anything, because the series you've called A and S are divergent. It doesn't make any sense to add and subtract divergent series, even if the result you appear to get is finite. I know it's tempting, but if you think about what you're doing, i.e. treating infinity as a number, then it's obvious that it's not admissible.

In fact, you have to be very careful even when dealing with infinite sequences that are convergent. You're OK if the series is "absolutely convergent", i.e. if the sum of the absolute values of the terms in the series is convergent, but otherwise (e.g. the series for ln 2 that you've quoted above), all sorts of nasty things can go wrong if you change the order of the terms.

In fact, the first series you''ve written down is quite interesting. In a sense, it is the slowest-diverging series that you can get. That's a bit vague (and is possibly not even true), but consider the following function

f(z) = S¥ n=1n^-z

where z is a complex number (if you're not familiar with complex numbers, you can take it to be a real number - it doesn't matter too much). Notice that f(1) is just your S, which is divergent. But it turns out that for ANY z (whose real part is) strictly greater than 1, the series converges. So you can make z as close to 1 as you like, so long as it is bigger than 1, and the series f(z) will converge. That's what I mean when I say it converges "as slowly as possible".

f(z) is called the Riemann zeta function, which you may have heard of. It turns out that there is a natural way of extending its domain of definition (i.e. the set of complex numbers z for which it exists) to all the complex numbers, except for z=1. The most famous unsolved problem in mathematics, the Riemann hypothesis, concerns the set of values of z for which f(z) = 0: are there any whose real part is not equal to 1/2? Mathematicians a pretty sure that there are not, and they've done a lot of work based on that assumption, but no one knows for sure....

Forgive me if I've wandered too far from the point. The first paragraph answers your question, I hope. If you go on to study maths at degree level, you will be taught how to deal rigorously with infinite series, and be taught techniques for determining which series converge and which diverge. Incidentally, can you prove that S diverges?

By Andrew Smith (P2517) on Wednesday, May 24, 2000 - 09:54 am:

Thank you both for your replies, they have cleared up a lot of confusion.

Coincidentally I was reading about the Riemann Zeta function a few days ago in Jan Gullberg's Mathematics, From the Birth of Numbers, a sort of history of mathematics so your (Simon) reply did not really wander.

The book says,
"The function f(z) has no zeros in Re(z) greater than or equal to 1; its only zeros in Re(z) less than or equal to 0 are at z=-2,-4,-6,...; it has infinitely many zeros in 0<Re(z)<1, which are called non trivial zeros."
I understood zeros to mean values of z where f(z)=0 (is this correct?)
but if f(z) = 1 + 1/2z + 1/3z +etc.
f(-2) = 1 + 1/2-2 + 1/3-2 = 1 + 4 + 9 + ... which isn't equal to zero, where am I going wrong?

Also the book says,
"it has been shown that the first 1.5×109 zeros in 0<Re(z)<1 are all nontrivial zeros on Re(z)=1/2"
How are the zeros ordered? Is it by magnitude of z?

Can you please either (a) give me answers to these questions or if you are busy (b)point me in the direction of a good book or internet site.

Sorry to be asking so many questions near exam time but I need something to relieve the tedium of A-level revision! Thanks again

By Simon Munday (Sjm78) on Wednesday, May 24, 2000 - 06:10 pm:

Hi Andrew,

I think I can answer your questions. Firstly, you're right about what a zero is. Secondly, notice that you can't use the series definition of the zeta function for values of z less than 1, because it doesn't converge (you're still being tempted to accept divergent series! It's not allowed!) Instead, there is another definition of the function (in terms of a contour integral and something called the gamma function - don't worry about what they are) which works for every value of z (apart from z=1) and which is exactly equal to the series definition for values of z for which the series converges. So when the book says that f(-2) = 0, it doesn't mean that the sum 1 + 4 + 9 + ... = 0, it means that the other definition has a zero at -2. This integral definition of f is called the analytic continuation of f into the subset Re(z) <= 1 of the complex plane. It turns out that in general, analytic continuations are unique - there is at most one way of extending a function defined on a limited subset of C to a larger subset such that the extended function is analytic (which means, roughly, that it is complex differentiable).

I'm not sure how non-trivial zeros are ordered, to be honest. It's probably by the size of the imaginary part, seeing as they all have the same imaginary part. I doubt it's particularly important, though.

I hope that helps. The moral is, if a series diverges, you can't pretend it doesn't.
You'll be found out in the end (generally, by producing absurd results).

By Dan Goodman (Dfmg2) on Wednesday, May 24, 2000 - 07:03 pm:

It's interesting to note that the mathematicians before the 1850s or thereabouts (the dates may be wrong) didn't understand these things. Many famous (and great) mathematicians made just this sort of mistake. Sometimes the results they came up with were OK, sometimes not. Euler apparently once wrote that 1+2+4+8+...=-1, because if you take x=1+2+4+8+..., multiply by 2 and add 1, you get 2x+1=1+2+4+8+..., so 2x+1=x, so x=-1.

By Andrew Smith (P2517) on Thursday, May 25, 2000 - 10:23 pm:

Another error attributed to Euler,
..... x-3 + x-2 + x-1 + x0 + x1 + x2 + x3 ...... (extending infinitely in both directions)
= (..... x-3 + x-2 + x-1) + (x0 + x1 + x2 + x3 ......)
= 1/(x-1) + 1/(1-x)
= 0.

Thanks again for the emails, in my defence I wasn't trying to say that 1+4+9...=0, I just wanted to know how it worked. The book I was reading literally had one page about this, the formula and a few facts.

You asked if I could show that the harmonic series was divergent, apart from the standard 2n pieces proof I found another today. In Mathematical Morsels by Ross Honsberger there is the problem; prove that 1/n + 1/(n+1) +...+ 1/(n2-1) + 1/n2 > 1.
The proof given is 1/n + 1/(n+1) +...+ 1/(n2-1) + 1/n2 > 1/n + (1/n2)(n2-n)
= 1/n + 1 - 1/n = 1.
If we set n=1 then n=2 then n=5 then n=26 where each subsequent n is equal to the previous n squared plus one then all the recipricals are covered and each new n adds one to the total and so the sequence diverges. It's not as elegant as the standard proof but it's something different I suppose.

On another note if the fibonacci numbers start f1=1, f2=1, it appears that if;
a prime p is of the form 10n+1 or 10n-1 it divides fp-1
a prime p is of the form 10n+3 or 10n-3 it divides fp+1
a prime p is of the form 5n then it divides fp+1 (kind of trivial)

This works for p up to about 127 (as far as I've checked) unless I've messed things up.
Does anyone know of a proof of this or counterexample?

Cheers

By David Seery (Djs61) on Friday, May 26, 2000 - 01:51 am:

I may be getting myself into trouble here, because it's not a subject I know much about, but it is said that the odd results 1+1+1+... = -1/2 and 1+(1/2)+(1/4)+... = -1/12 (I think those are right - they come from the zeta function contour integral in a dubious kind of way) are used in Quantum Field Theory to renormalise these expressions. Apparently the answers do check with reality. I'm not really sure about this, though ...

By Sean Hartnoll (Sah40) on Friday, May 26, 2000 - 12:01 pm:

As far as I understand (and I haven't done a course in QFT yet), quantum field theory doesn't claim that these sums are true. What is happening is that an infinite amount of energy is substracted off. This is (sort of) justified on the grounds that the origin of energy is unimportant as only changes in energy matter.

More technically, what happens is that in the classical theory, you have variables that commute, but when you go over to quantum mechanics they don't. Now depending which order you take the variables in, you get different energies, sometimes infinite. You then do a reordering in the classical theory, which is allowed, so that when you go over to quantum theory the results comes out finite. Obviously this is dodgy. And it does't work for gravity. But QFT is the most accurate theory around.

Sean