>> There's no grouping like you did in the original proof. Which part of
>> the original proof assumes the length of the summation is anything other
>> than infinite?
>
> This part of the proof is grouping elements in pairs and summing them up:
>
>    (s-4s) = 1+2+3+4+5+ 6+...
>               -4  -8  -12-...
>    -3s    = 1-2+3-4+5-6+...

This one is different, as there is no assumption that the "..." 
(infinite list) has any other properties other than "it goes on 
forever". So long as both parts of the sum "go on forever" then there 
will always a pair for each item.

However if you try and group elements like:

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

Then the "..." in the 3rd (and perhaps 2nd) line makes the assumption 
that there are an even number of terms, that the series ends in a "-1". 
That (I think) is a wrong assumption, an infinite list/sum doesn't have 
any concept of a "last" item.

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