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Re: It is a very bad idea and nothing less than stupid to define 1/3 = 0.333...
Posted:
Oct 3, 2017 5:33 PM


Why are they the same? Simply because oo is not an element from the natural numbers. So
\sum_{k=1}^oo 9/(10^k)
cannot mean something where the last summand is 9/(10^oo) because a value 9/(10^oo) doesn't exist
in Q, since there is no k=oo. So all you have is the partial sums that go on and on,
\sum_{k=1}^n 9/(10^k) 0.9 0.99 0.999 ...
And you cannot identify a final value. To extract a value from the series you need the limes.
Am Dienstag, 3. Oktober 2017 23:21:40 UTC+2 schrieb burs...@gmail.com: > Since \sum_{k=1}^n 9/(10^k) = 1  10^(n), > and \sum_{k=1}^oo a(k) = lim n>oo \sum_{k=1}^n a(k) > > The two are synonymous: > > lim n>oo (110^(n)) = \sum_{k=1}^{\infty} 9/(10^k) > > https://en.wikipedia.org/wiki/Synonym > > BTW: The following authors here on sci.math already > explained this two you like a dozen times: >  Dan >  Me >  Markus Klyver >  Zelos Malum >  Etc.. > > You, bird brain John Gabriel, probably qualify > for the most stupid human being on the planet. > > Should we call the Guiness book of records? > > Am Dienstag, 3. Oktober 2017 22:04:52 UTC+2 schrieb John Gabriel: > > > lim n>oo (110^(n)) > > > > No. 0.999... = \sum_{k=1}^{\infty} 9/(10^k)



