1 + 10 + 100 + 1000 + = -1/9

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[tex]S = 1 + 10 + 100 + 1000 + 10000 + ... [/tex]

[tex]10S = 10 + 100 + 1000 + 10000 + 100000 + ... [/tex]

[tex]S - 10S = (1 + 10 + 100 + 1000 + 10000 + ...) - (10 + 100 + 1000 + 10000 + ...) [/tex]

[tex]-9S = 1 + (10 - 10) + (100 - 100) + (1000 - 1000) + (10000 - 10000) ... [/tex]

[tex]-9S = 1 + 0 + 0 + 0 + 0 + 0 ...[/tex]

[tex]-9S = 1 [/tex]

[tex]S = -1/9[/tex]

:confused:
What's wrong (or right) with this?

Thanks,
Unit
 

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  • #4
Hurkyl
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Also, I believe that sum converges as an ordinary infinite sum in the 2-adics and the 5-adics.

(And, of course, it does not converge as an ordinary infinite sum in the reals!)
 
  • #5
CRGreathouse
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I would have [intuitively] expected it to converge in all the p-adics. Am I wrong?
 
  • #6
Hurkyl
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In any other p-adic field, the terms don't converge to zero!
 
  • #7
CRGreathouse
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In any other p-adic field, the terms don't converge to zero!
:blushing:
 
  • #8
Char. Limit
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I'm pretty sure that you can't pair up terms in an infinite sum.
 
  • #9
Redbelly98
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[tex]S = 1 + 10 + 100 + 1000 + 10000 + ... [/tex]
.
.
.
What's wrong (or right) with this?

Thanks,
Unit
I'm pretty sure that you can't pair up terms in an infinite sum.
The way I remember it, proofs like this actually say something like:
If S exists, then S = 1 + 10 + 100 + ...​
So if S does not exist, then the remaining statements do not necessarily hold true.
 
  • #10
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If S exists, then S = 1 + 10 + 100 + ...​
So if S does not exist, then the remaining statements do not necessarily hold true.
:smile: Brilliant! I had completely forgotten about variables and their related hypothetical syllogisms. Thanks!
 

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