I think the problem is that for non-converging series, there are multiple
similar tricks you could do that would give different answers...for example:

S = 1-1+1-1+1-1...
-1*S = -1+1-1+1-1+1...

For a finite or converging series, the order of the summation doesn't
affect the final sum, so if in -1*S we switch the places of the 1st and 2nd
term, similarly with the 3rd and 4th, etc., we get:

-1*S = 1-1+1-1+1-1... = S

But if S = -S this implies 2S = 0 which means S=0.

Another way of getting S=0 would just be to group terms together like this:

S=1-1+1-1+1-1...=(1-1)+(1-1)+(1-1)+...=0+0+0..., which is a convergent
series with a well-defined sum of 0.

Likewise, you could also group the terms together like this:

S = 1-1+1-1+1...=1+(-1+1)+(-1+1)+(-1+1)+...=1+0+0+0..., which again is a
convergent series with a sum of 1.


On Wed, Jan 29, 2014 at 3:56 PM, LizR <[email protected]> wrote:

> OK... thanks, I should have guesses it was the zeta function :D
>
> Anyway, I showed this proof to my 15 year old son and he soon put me right
> on why 1-1+1-1+1-1+1... is indeed 1/2.
>
> call the series 1-1+1-1+1... S
>
> then 1-S = 1 - (1-1+1-1+1-1+1...) = 1-1+1-1+1-1... = S
>
> S=1-S, so S=1/2 (which is, I should think, another way of writing Bruno's
> proof, above, but maybe even simpler!)
>
> Actually that *does *look rigorous. I mean, assuming that infinite series
> exist and can be added up, etc, etc, that answer looks fairly watertight.
> What could possibly go wrong?
>
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