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Re: .99999... still=/= 1
Posted:
Dec 1, 2004 4:44 PM
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smart1234@aol.com (S. Enterprize Company) wrote in
<a href="news://20041201034756.21909.00001534@mb-m06.aol.com:">news://20041201034756.21909.00001534@mb-m06.aol.com:</a>
>>smart1234@aol.com (S. Enterprize Company) wrote in
>><a href="news://20041130053834.21841.00001212@mb-m14.aol.com:">news://20041130053834.21841.00001212@mb-m14.aol.com:</a>
>>
>>> You could use the calculator to see if this diverges. If the
>>> calculator say
>>> number too large, then you know it diverged.
>>>
>>
>>Try this one on your calculator,
>>
>>(10^2000000) + (10^2000000)/2 + (10^2000000)/4 + (10^2000000)/8 +
>> + (10^2000000)/16 + (10^2000000)/32 + ...
>>
>>Tell me if your calculator says it "diverges." Then I will tell you
>>want good ol' logic says happens.
>
> It diverges, I didn't need the calculator.
>
> That series is of the form,
>
> 1 + 1/2 + 1/4 + 1/8+ ...
>
> This diverges.
>
>
No, it converges. Factor out the 10^2000000 and you have
(10^2000000)*(1 + 1/2 + 1/4 + 1/8 + ...)
=(10^2000000)*(2)
=(2*10^2000000)
Now, I am willing to bet your calculator will give you a "really big
number." and by your method of determining convergence or divergence on a
calculator, it would diverge.
Clearly that is wrong. It converges to a number that is just too large for
you calculator, but a very finite number.
- Tim
--
Timothy M. Brauch
NSF Fellow
Department of Mathematics
University of Louisville
email is:
news (dot) post (at) tbrauch (dot) com
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