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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: Replacement of Cardinality
Date: Tue, 27 Aug 2024 13:14:54 -0700
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On 8/27/2024 1:01 PM, Moebius wrote:
> 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) = ?

3/4

Yup. Never equals one... However it "tends" to it... Yet the sum will 
never be one... :^)


> 
> 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.

I think so. :^)