Logarithm reciprocal

[Gottfried]

Code:

14879 1.44270934056096
29759 1.44270932083314

  oo  1.44269504088896?

Hmm, I got no better result yet, but maybe you see something further.

I reformulated this into

\(
\lim_{n->\infty} (n+1) \sum_{k=1}^{\infty} \( 1-\frac1{2^k} \)^n*\frac1{2^k} =^{?}_{ } \frac1{\ln\(2\)} \)

or

\(
\lim_{n->\infty} \sum_{k=1}^{\infty} \( 1-\frac1{2^k} \)^n*\frac1{2^k} =^{?}_{ } \frac1{\ln\(2^{n+1}\)}
\)

to remove the binomials first. So far I get numerically the same results as in your examples (first version of my equation).
Though I don't see much more clearer here.

Perhaps we can relate this to Euler's "false logarithmic series" which provides a polynomial approximation to the log which is only valid at the integers (but not at zero(!)) and has a sinusoidal deviation from the true log-function - but I didn't go deeper into this yet...

Gottfried