>> -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.

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