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Han de Bruijn wrote:
> You can repeat this a thousand times, but Tony and I don't get it. In my
No kidding....
> not so humble opinion, you must also reject then the integral(0,1) dx ,
> because it is derived from the Riemann sum n.1/n , which is exactly the
> same as summing up (n) probabilities with 1/n chance for each. dx = 1/n.
> Now take the limit for n->oo and you're done. What's the problem?
There is no problem. Nobody disputes that
lim_{n \rightarrow \infty} (n. 1/n) = 1
But nobody except you (and Tony?) thinks that you can meaningfully
claim
that
lim_{n \rightarrow \infty} (n.1/n) =
(lim_{n \rightarrow \infty} n).(lim_{n \rightarrow \infty} 1/n)
The result lim(a_n b_n) = (lim a_n)(lim b_n) is only guaranteed to
be true if a_n and b_n both converge. Since this condition is not
satisfied in the above example, there is no foundation for your
argument.
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