Re: Infinite multiplication
 

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I wrote a perl script (before I saw your anwer), and it didn't seem to converge.

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This shows that it is converging. Convergence means that it gets closer and closer to some limit, even if it never actually gets there, and even if we can't identify the limit by any means other than by continuing the calculation to greater precision, as with the calculation of pi. Obviously if it converges to an irrational number, your calculations will never terminate. We know that this converges because the steps in the sum of logs are getting smaller slightly faster than the geometric series, which we know converges.

Re: Infinite multiplication
 

If there is some fixed k<1, such that every term in your sequence is less than k, that is sufficient to show convergence to 0. The n_th partial product will be less than k^n and the power series with k<1 converges to 0.

For instance, 0.9 x 0.95 x 0.955 x 0.9555 x 0.95555 ... will converge to zero, because every term is less than 0.96.

Sufficient, but not necessary - the terms can get arbitrarily close to 1 provided they do so very slowly (more slowly than e^(-1/n) approaches 1, I think.)

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.

Re: Infinite multiplication
 

Thanks. I'm amazed after not doing pure mathmatics for over 20 years how the beauty of it still amazes me.

Re: Infinite multiplication
 

Thanks for all the replies in this thread.