Ivars Wrote:next ln2+ ln*lnGamma(1/2) - lnln2 etc. ln2 and lnln2 will cancel out, so
in the end we will have lnln............ln (Gamma(1/2) when n-> infinity.
Actually it does not matter with which value (instead of -1) you start the iterations of \( \log_b \) as long as you dont take some hyperpowers of \( b \) for example \( b^{b^b} \) would lead to 0 after 4 times application of the \( \log_b \) and taking again \( \log_b \) would give \( -\infty \), so the formula would not work with this starting value.
