Re: Replacement of Cardinality

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. :^)