12/30/2007, 01:09 PM

One application of h(1/(n^n)) = 1/n as n-> infinity

I will try to apply my infinite tetration formula h(1/(n^n)) = 1/n to see what happens:

So as n-> infinity

(e^n)(n!)/((n^n)(sqrt n))=sqrt 2pi (Clawson)

1/(n^n) = (sgrt(2pi*n)) / ((e^n)*n!)

h(1/(n^n)) = h( sgrt(2pi*n)/(e^n)*n!) = 1/n as n-> infinity which is? 0?

If I knew how h works on composite arguments, may be it would be possible to study this deaper.

I will try to apply my infinite tetration formula h(1/(n^n)) = 1/n to see what happens:

So as n-> infinity

(e^n)(n!)/((n^n)(sqrt n))=sqrt 2pi (Clawson)

1/(n^n) = (sgrt(2pi*n)) / ((e^n)*n!)

h(1/(n^n)) = h( sgrt(2pi*n)/(e^n)*n!) = 1/n as n-> infinity which is? 0?

If I knew how h works on composite arguments, may be it would be possible to study this deaper.