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logarithm reciprocal limit
Posted:
Jul 20, 2010 3:00 AM


In my current research I encountered the following intriguing sequence:
Define a_n recursively in the following way (for any b>0):
a_1 = 1 and for n>1: a_n = 1/(1b^n) * sum from m=1 to n1 over a_m * (n over m) * (1b)^(n m) * b^n
For ASCII handicapped people here the typeset formula: http://math.eretrandre.org/cgibin/mimetex.cgi?a_n%20=%20\frac{1}{1b^n}\sum_{m=1}^{n1}%20a_m%20\left(n\\m\right)%20(1b)^{nm}%20b^m
My conjecture is that a_n > (b1)/ln(b)
Have anyone heard about such a formula, or can prove or disprove its convergence? It sounds too elementary for not being treated somewhere already.



