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Old 12-09-2005, 04:22 AM
jason_t jason_t is offline
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Join Date: Nov 2004
Location: Another downswing?
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Default Re: Infinite multiplication

If a_n is a sequence of complex numbers and sum |a_n| converges then product (1 + a_n) exists and is zero iff a_n = -1 for some n. Furthermore, the convergence of this product is unconditional in that for any permutation n_1, n_2, n_3, ... of 1, 2, 3,... product (1 + a_{n_k}) converges and equals product (1 + a_n). If a_n = a_n(s) depends on a parameter s sum that sum |a_n(s)| converges uniformly in s then the product product (1 + a_n(s)) converges uniformly on S too and the product converges to zero at s_0 in S iff a_n(s_0) = -1 for some n. The convergence is unconditional on S. This is a powerful theorem and its proof is based on the trivial inequality 1 + x <= e^x for x >= 0.
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