05/15/2014, 08:53 PM
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
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

