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Topic: logarithm reciprocal limit
Replies: 8   Last Post: Aug 11, 2010 7:30 AM

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Posts: 83
Registered: 6/11/06
Re: logarithm reciprocal limit
Posted: Jul 23, 2010 4:00 PM
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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:

Your formula $\frac{1-(1-b)^n - nb(1-b)^{n-1}}{1-b^n}$ 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,...
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 non-recursive simple formula:

a_n = (b-1)\sum_{k=1}^{n} \binom{n}{k} \frac{k (-1)^k}{1-b^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}{1-b^k}
tends to 1/\ln(b)?

Here is picture of the numerical behavior of b_n for b=2:
It oscillates with decreasing amplitude and increasing period around
1/ln(2)=1.442695... (the y-axis numbers are however messed up by Sage)

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