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From: Moebius <invalid@example.invalid>
Newsgroups: sci.math
Subject: Re: Replacement of Cardinality
Date: Tue, 27 Aug 2024 22:01:15 +0200
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Am 27.08.2024 um 21:34 schrieb Chris M. Thomasson:
> On 8/27/2024 12:20 PM, WM wrote:
>> Le 25/08/2024 à 23:18, Richard Damon a écrit :
>>>
>>> But the sum [...] adding up the terms of
>>>
>>> 1/n - 1/(n+1)
>>>
>>> never gets to 1,
> 
> 1/1 - 1/2 = .5
> 1/2 - 1/3 = .1(6)
> 1/3 - 1/4 = .08(3)

Don't forget to sum up the terms:

     (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) = ?

If we look closely we see that most of the terms (...1/k...) can be 
ignored. We just have to consider:

     1/1 - 1/4 = 3/4.

Check: .5 + .1666... + .08333... = 0.75 = 3/4. (ok)

Moreover we can prove this way that (as claimed above)

     SUM_(k=1..n) 1/k - 1/(k+1) < 1   (for all n e IN).

Proof: We consider the sum of the terms for k = 1..n (where n is an 
arbitrary natural number): (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... 
+ (1/(n-1) - 1/n) + (1/n - 1/(n+1)) = 1/1 - 1/(n+1) = 1 - d, where d > 
0. Hence: SUM_(k=1..n) 1/k - 1/(k+1) < 1. qed

>> It does, but only in the darkness.
> 
> Are you a full blown moron or just _mostly_ stupid?

I'd say BOTH.