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But the one where we want the product to converge has it converging to something nonzero.
by NotYourLawyer 3y ago
But the one where we want the product to converge has it converging to something nonzero.
- scterg 3y agoSee the wiki link in a previous comment: https://en.wikipedia.org/wiki/Infinite_product https://en.wikipedia.org/wiki/Infinite_product "The product is said to converge when the limit exists and is not zero." #1. Consider the series 1 - 1/2^(1/2) + 1/3^(1/2) - 1/4^(1/2) + ... . The sum converges, but the infinite product diverges to zero. #2. Consider the series x_1 - y_1 + x_2 - y_2 + x_3 - y_3 + ... where x_n and y_n satisfy x_n - y_n = 1/n, x_n * y_n = 1/n. It is obvious that the sum is the harmonic series and diverges. However, the product converges to one, seeing as x_n \to 0 and (1 + x_n) * (1 - y_n) = 1.
- NotYourLawyer 3y agoI like it. I didn’t realize that “diverging to zero” was a thing—wonder if that’s what the author had in mind, or if there’s an example with real divergence.
- scterg 3y agoIndeed. I didn't realize "diverging to zero" with a thing either. I do think this is what the author had in mind, since he writes about the sum of the logs, "whose convergence is equivalent to the convergence of the product".
- NotYourLawyer 3y agoYeah, but that’s in the context of the theorem that requires the terms to be greater than zero.