Tricky Infinite Series and Radius of ConvergenceDate: 04/21/2008 at 21:32:09 From: Adam Subject: A Very Tricky Series (Radius of Convergence-Endpoint Test) This problem has consumed and befuddled 5 professors, and the entire Tutor Center (me included) at Sierra College here. Any help on this final portion of the problem would be greatly helpful (considering this is all I've been trying to solve for the past week). This has no grade value or anything of the sort, its just a "What-If" situation. I will not show the original problem because there is only the issue of the limit. But an infinite series is given, and a radius of convergence is determined with endpoints (8/3) and (16/3). The left endpoint is proven to be convergent, but as much as we extrapolate the right endpoint (16/3), we can only assume it's divergent, without any solid proof. Here's the series in question (I apologize about the poor notation, I don't have MathType handy): SUM 'n=1->inf' (4/3)^n * [n!(1*4*7*10...{3n-2})] / [(2n)!] I can't really write out all the work for this, but I can list some of what has been attempted: Ratio/Root Tests fail with "Indeterminate" No way on the Integral Test Simple Divergence Test is redundant-gets us nowhere It's not an Alternating Test The most appealing was the Limit Comparison Test (such as relating to (1/n), which is divergent. We can, via Excel, show massive divergence at 100 terms, but how can we algebraically show it? I do not expect a detailed answer, just a shove in the right direction if possible. Thank you. Date: 04/22/2008 at 20:00:47 From: Doctor Vogler Subject: Re: A Very Tricky Series (Radius of Convergence-Endpoint Test) Hi Adam, Thanks for writing to Dr. Math. The first thing that comes to mind is to use Stirling's Approximation http://mathforum.org/library/drmath/view/55996.html to approximate the two factorials. Of course, the product with every third integer isn't as nice, so we have to handle this separately. Well, one way to get something almost as good as Stirling's approximation is to use the idea of Formula to Sum a Series of Square Roots http://mathforum.org/library/drmath/view/65309.html We can use this to prove that S = SUM 'k=1->inf' log(3k-2) has -log 2 <= S - [(n - 1/6)*log(3n - 2)] - n + 1] <= 0, which means that (1*4*7*10...{3n-2}) is roughly (to within a multiple of 2) (3n - 2)^(n - 1/6) * e^(-n + 1) Combining this with Stirling's approximation (which says that n! is about n^(n + 1/2) * e^-n), we find that your summand (4/3)^n * [n!(1*4*7*10...{3n-2})] / [(2n)!] is roughly (to within a constant multiple) (4/3)^n * n^(n + 1/2) * e^-n * (3n - 2)^(n - 1/6) * e^(-n + 1) * (2n)^(-2n - 1/2) * e^(2n) = (4/3)^n * n^(n + 1/2) * (3n - 2)^(n - 1/6) * (2n)^(-2n - 1/2) * e = (1/3)^n * n^(-n) * (3n - 2)^(n - 1/6) * e/sqrt(2) = (1/(3n))^n * (3n - 2)^(n - 1/6) * e/sqrt(2) = ((3n - 2)/(3n))^n * (3n - 2)^(-1/6) * e/sqrt(2) = (1 - (2/3n))^n * (3n - 2)^(-1/6) * e/sqrt(2) I'll leave it to you to verify that the limit as n goes to infinity of (1 - (2/3n))^n is e^(-2/3), and more generally that for n >= 1, 1/3 <= (1 - (2/3n))^n < e^(-2/3), which means that your summand is always within a constant multiple of (3n - 2)^(-1/6) and now you can use the p-test, or the integral test, to verify that this sequence is divergent. In other words, you have (4/3)^n * [n!(1*4*7*10...{3n-2})] / [(2n)!] > C*(3n - 2)^(-1/6) for some positive constant C (such as C = 1/6), and the series on the right is divergent. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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