Searching for an asymptotic to exp[0.5]
#19
To justify a bit more what I said notice that :

Let D be the differential operator.

\( \lim_{n->\infty} \frac{H_n}{\ln^{[1/2]}(n) } -> 1 \)

This follows from

\( \lim_{n->\infty} \frac{D \exp^{[1/2]}(n)}{\exp^{[1/2]}(n) } -> 1 \)

Now D exp^[1/2](x) is finally smaller than A * exp^[1/2](x) for any A > 1 BECAUSE

exp^[1/2](x) < exp(A x)

Take the logarithmic derivative on both sides :

( D exp^[1/2](x) ) / exp^[1/2](x) < A

QED.

Ofcourse we know D exp^[1/2](x) > exp^[1/2](x) / x from the simple consideration of a Taylor series.

This proves the previous post of me was correct.


regards

tommy1729
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Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 05/15/2014, 08:53 PM

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