Thread: Problem 37
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Old September 17th, 2007, 08:40 PM
putnam120 putnam120 is offline
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#1) the initial values of a_0 and a_1 do not matter. you just need to notice that \lim_{n\to\infty}\frac{a_n}{a_{n-1}}=\phi=\frac{1+\sqrt{5}}{2}
so to find the radius we just use the ratio test and get

\lim_{n\to\infty}\left|\frac{x^na_n}{x^{n-1}a_{n-1}}\right|=\lim_{n\to\infty}\left|x\left(1+\frac{a_{n-2}}{a_{n-1}}\right)\right|=|x|\left(1+\frac{1}{\phi}\right)<1

so |x|<\frac{1}{1+\frac{1}{\phi}} which makes the radius of convergence \frac{1}{1+\frac{1}{\phi}}=\frac{\phi}{1+\phi}
Last edited by putnam120; September 17th, 2007 at 09:48 PM.