On Wed, Jan 29, 2014 at 9:56 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?
I've noticed something (maybe silly, maybe trivial?). Let's say: S(0) = 1 = 1 S(1) = 1 - 1 = 0 S(2) = 1 - 1 + 1 = 1 S(3) = 1 - 1 + 1 - 1 = 0 S(inf) = 1/2 So the summation oscillates between 0 and 1, and at the limit it's in the middle of these two values. Notices that for 2 - 2 + 2 - 2 + 2... the summation oscillates between 0 and 2 and it's 1 at the limit, and so on. > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To post to this group, send email to [email protected]. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/groups/opt_out. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To post to this group, send email to [email protected]. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
