tag:blogger.com,1999:blog-12604614.post116306583939939838..comments2023-06-12T05:26:38.585-07:00Comments on Math Refresher: Absolute Convergence for Infinite ProductsLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-12604614.post-52559272888461314092007-04-16T18:01:00.000-07:002007-04-16T18:01:00.000-07:00Anonymous,Thanks very much for noticing the typos....Anonymous,<BR/><BR/>Thanks very much for noticing the typos. I have updated the lemmas that you mention.<BR/><BR/>In reviewing it, I realize that these lemmas are far from being as clear as they should be.<BR/><BR/>When I get a chance, I plan to simplify them. Please continue to let me know if you notice any more inaccuracies.<BR/><BR/>Thanks very much,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12604614.post-24240114548424285212007-04-03T02:44:00.000-07:002007-04-03T02:44:00.000-07:00Lemma 1 is wrong, lemma 4 is wrong for your a,i gr...Lemma 1 is wrong, lemma 4 is wrong for your a,i greater or equal to zero.<BR/><BR/>Lemma 1 is true for |x|<1. Thus ur lemma 4 is correct for the terms <BR/>-1 < a,i <0.<BR/><BR/>Denote a,i as a subscript i<BR/><BR/>For a,i>=0, use this ln (1+|a,i|) = ln (1+a,i) <= a,i = |a,i|<BR/>which follows easily from ln (1+x)<= x, which follows from e^x>= 1+xAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-12604614.post-43779970509834458572007-02-15T11:11:00.000-08:002007-02-15T11:11:00.000-08:00Thanks very much for your comment.I agree with you...Thanks very much for your comment.<BR/><BR/>I agree with you that I should make it clear that abs(x) ≠ 1.<BR/><BR/>I've updated the lemma.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12604614.post-36541439030329235602007-02-15T09:37:00.000-08:002007-02-15T09:37:00.000-08:00lemma 1 is not true for x>1: the series 1+x+x^2+.....lemma 1 is not true for x>1: the series 1+x+x^2+... has a singularity in x=1, and so cannot be extended from 0 to infinityAnonymousnoreply@blogger.com