
Re: Finally the discussion is over: S = Lim S is a bad definition.
Posted:
Oct 3, 2017 9:01 PM


But if you plot the two series, 0.999... and 0.999...9995 you will see that neither of them is below the other, there is always one summand that makes it bigger:
0.999... 0.999...9995 0.9 0.95 0.99 0.995 0.999 0.9995 Etc..
So we can only conclude Eulers series are brilliant, interpreting a1+a2+a3+... as lim n>oo sum an, is the only viable way,
any nonstandard numbers with 0.999... != 1 dont make any sense, they dont obey the standard laws of algebra and blow up the
number space with a lot of cripples.
Am Mittwoch, 4. Oktober 2017 02:32:47 UTC+2 schrieb burs...@gmail.com: > Well its the mean value between 0.999... and 1.0. > Look if we have a and b, the mean value is (a+b)/2: > > a b (a+b)/2 > 0.9 1.0 0.95 > 0.99 1.00 0.995 > 0.999 1.000 0.9995 > 0.9999 1.0000 0.99995 > Etc.. > > If 0.999... <> 1, then also 0.999... <> 0.999...9995 > and then also 0.999...9995 <> 1. > > And so on, as soon 0.999... <> 1, there are miriad > other numbers inbetween. > > My suggestion: Don't do this nonsense, just interpret > 0.999... as limit. Then you can also interpret > > 0.999..9995 as limit, namely this charming limit, > the following summands summed up to finity: > > 0.5 > 0.45 > 0.045 > 0.0045 > ... > > Guess what is the result? > > 1/2+sum_i=1^n (45/10^(i+1)) = 1/2 + (1  10^(n))/2 > > lim n>oo 1/2 + (1  10^(n))/2 = 1 > > Am Mittwoch, 4. Oktober 2017 02:24:32 UTC+2 schrieb burs...@gmail.com: > > So whats this Gabriel number: > > > > lim n>oo 0.999...9995 = ? > > \n/ > > > > Am Mittwoch, 4. Oktober 2017 02:08:09 UTC+2 schrieb burs...@gmail.com: > > > So if 0.999... <> 1, do we have: > > > > > > 0.999... < 1 > > > > > > Or rather? > > > > > > 0.999... > 1 > > > > > > Am Mittwoch, 4. Oktober 2017 01:40:31 UTC+2 schrieb John Gabriel: > > > > > nonsense. Right. But this was clear from the beginning. > > > > > > > > Yes. Of course S = Lim S is nonsense. So why do you still believe in it?

