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Topic: .99999... still=/= 1
Replies: 43   Last Post: Dec 8, 2004 3:31 PM

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 Tim Brauch Posts: 201 Registered: 12/6/04
Re: .99999... still=/= 1
Posted: Dec 1, 2004 4:44 PM

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>

&gt;&gt;smart1234@aol.com (S. Enterprize Company) wrote in
&gt;&gt;<a href="news://20041130053834.21841.00001212@mb-m14.aol.com:">news://20041130053834.21841.00001212@mb-m14.aol.com:</a>
&gt;&gt;
&gt;&gt;&gt; You could use the calculator to see if this diverges. If the
&gt;&gt;&gt; calculator say
&gt;&gt;&gt; number too large, then you know it diverged.
&gt;&gt;&gt;
&gt;&gt;
&gt;&gt;Try this one on your calculator,
&gt;&gt;
&gt;&gt;(10^2000000) + (10^2000000)/2 + (10^2000000)/4 + (10^2000000)/8 +
&gt;&gt; + (10^2000000)/16 + (10^2000000)/32 + ...
&gt;&gt;
&gt;&gt;Tell me if your calculator says it "diverges." Then I will tell you
&gt;&gt;want good ol' logic says happens.
&gt;
&gt; It diverges, I didn't need the calculator.
&gt;
&gt; That series is of the form,
&gt;
&gt; 1 + 1/2 + 1/4 + 1/8+ ...
&gt;
&gt; This diverges.
&gt;
&gt;

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