Logarithm reciprocal

[bo198214]

I reformulated this into

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

Yes, expanding \( \frac{1}{1-b^k}=\sum_{i=0}^\infty b^{ki} \) and then applying the binomial formula. But I was still unable to make something nice out of this formula.

I was now going one step further and multiplied the sequence \( a_n \) with \( l_n = -\sum_{k=0}^n \frac{(1-b)^k}{k} \), i.e. the logarithmic series which we know must tend to \( \ln(b) \) (for \( 0<|1-b|<1 \)), so if \( a_n \) converges then \( a_n l_n \) should converge to 1.
The graph for b=2 (which is on the boundary of convergence) behaves very nice, the period is now exactly 2.
   
And this is the graph for argumetns 2n:
   

Perhaps we can relate this to Euler's "false logarithmic series"

which is that formula?