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Anybody figure out answers for the two exercises at the end? Nothing is jumping out at me.
by NotYourLawyer 3y ago
Anybody figure out answers for the two exercises at the end? Nothing is jumping out at me.
- deleted 3y ago[deleted]
- blahblah213423 3y agonegative numbers
- NotYourLawyer 3y agoYeah but which ones?
- scterg 3y agoNot 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.