scott <sco### [at] scottcom> wrote:
> >> -6s    = 1-1+1-1+1-1+1-...
> >
> > Which is equal to:
> >
> >   -6s = (1-1)+(1-1)+(1-1)+...
> >       = 0+0+0+0+... = 0

> I'm no mathematician, but to do that you must make the assumption that 
> there are an even number of terms in the infinite sum (ie every +1 has a 
> -1 to pair with it). You could have assumed an odd number of terms and 
> got a sum of 1 instead.

> Writing the sum equals 1 minus the sum seems to avoid the need to make 
> such an assumption.

An infinite sum can't have an "odd" or an "even" number of terms.

But you bring a good point. If you pair the elements differently, you get:

 -6s = 1+(-1+1)+(-1+1)+(-1+1)+...
     = 1+0+0+0+0+... = 1

Therefore s = -1/6.

In fact, you can get basically any integer value you want for -6s when
you group the elements appropriately.

This goes to show that when you are dealing with infinities, you can
"prove" anything you want.

I think the original "proof" is bogus.

The "proof" using Riemann's zeta function is also bogus in a sense.
Riemann's zeta function is the infinite sum of 1/n^s, but only for
values of s so that Real(s) > 1. (The sum would give 1+2+3+4+... when
s = -1, but the zeta function is not 1/n^s for values of s < 1. It's
something a lot more complicated.)

-- 
                                                          - Warp

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