回复：sequences

You could get a close formula of A_n. f(x) = (1+x/1)(1+x/2)..(1+x/n)

A_n = 1/3 * (f(1) + f(w) + f(w^2)), where w, w^2 are roots of x^2 + x + 1 = 0.

f(1) = n + 1.

w, w^2 are complex conjugates. You could estimate |A_n - 1/3(n+1)| = 1/3 |f(w) + f(w^2)|.

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• 回复：回复：sequences -蓝糊涂- ♂ (61 bytes) () 01/15/2010 postreply 08:37:04