
Re: logarithm reciprocal limit
Posted:
Jul 23, 2010 4:00 PM


On 07/22/2010 08:30 PM, Valentin Fadeev wrote: > I found some mistakes in my proof above, because i was a bit careless > with boundary terms. The corrected derivation in LaTex can be viewed > here: > > http://learn.open.ac.uk/mod/oublog/view.php?user=597675 >
Your formula $\frac{1(1b)^n  nb(1b)^{n1}}{1b^n}$ can not be true, it does not give the correct values; The values of a_n with b=2 are for n=1,2,... 1, 4/3, 10/7, 152/105, 314/217, 940/651, 5678/3937, 1447504/1003935,... yours: 0, 4/3, 4/7, 8/15, 8/31, 4/21, 12/127, 16/255, 16/511
I think you can not apply summation by parts if the v depends on two variables, namely n and m.
In the meantime I found a nonrecursive simple formula:
a_n = (b1)\sum_{k=1}^{n} \binom{n}{k} \frac{k (1)^k}{1b^k}
However it doesnt trigger anything familiar in my mind.
So, does anybody know why or whether b_n=\sum_{k=1}^n \binom{n}{k} \frac{k (1)^k}{1b^k} tends to 1/\ln(b)?
Here is picture of the numerical behavior of b_n for b=2: http://math.eretrandre.org/tetrationforum/attachment.php?aid=719 It oscillates with decreasing amplitude and increasing period around 1/ln(2)=1.442695... (the yaxis numbers are however messed up by Sage)

