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Not sure about #1. Is it impossible? Convergence of the sum of a_n implies a_n goes to zero, which means we can use the theorem, right? What am I missing? #2. C
by scterg 3y ago
Not sure about #1. Is it impossible? Convergence of the sum of a_n implies a_n goes to zero, which means we can use the theorem, right? What am I missing?
#2. Consider the sequence 1, -1/2, 1, -1/2, .... The sum diverges while the product "nearly" converges. A slight tweak will allow the product to converge while still letting the sum diverge. Essentially we are taking advantage of the convexity of the log function.
- scterg 3y agoRe: #1. Found an example using complex numbers. Still can't think of one using the reals though. EDIT: learned about "diverge to zero", where the product goes to zero but the series is still said to "diverge" (since the sum of logs diverges).
- NotYourLawyer 3y agoBut 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.
- NotYourLawyer 3y agoThe theorem is for positive numbers.