My Math Forum In need of help disk, series test, taylor, and power series

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 May 21st, 2012, 01:58 PM #1 Newbie   Joined: May 2012 Posts: 1 Thanks: 0 In need of help disk, series test, taylor, and power series I don't know where to start for these series test all sum as n goes to infinity, n=1 ((2n-1)/(3n+1))^n ((3n^3 -2n^2 + 4)/ (n^5 -n^3 +2) ln((3n)/(4n+5)) (n(-3)^n)/(4^(n-1))
 May 22nd, 2012, 07:58 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: In need of help disk, series test, taylor, and power se $\sum_{n=1}^{\infty}\,$$\frac{2n-1}{3n+1}$$^n$ Ratio test: $\lim_{n\to\infty}\,\frac{$$\frac{2(n\,+\,1)\,-\,1}{3(n\,+\,1)\,+\,1}$$^{n+1}}{$$\frac{2n\,-\,1}{3n\,+\,1}$$^n}$ $=\,\lim_{n\to\infty}\,\frac{$$\frac{2n\,+\,1}{3n\, +\,4}$$^{n+1}}{$$\frac{2n\,-\,1}{3n\,+\,1}$$^n}$ $=\,\lim_{n\to\infty}\,$$\(\frac{2n\,+\,1}{3n\,+\,4 }$$\frac{$$\frac{2n\,+\,1}{3n\,+\,4}$$^{n}}{$$\fra c{2n\,-\,1}{3n\,+\,1}$$^n}\)$ $=\,\lim_{n\to\infty}\,$$\(\frac{2n\,+\,1}{3n\,+\,4 }$$$$\frac{(2n\,+\,1)(3n\,+\,1)}{(3n\,+\,4)(2n\,-\,1)}$$^{n}\)$ $=\,\frac23\,\cdot\,1\,=\,\frac23$ Thus, the series converges absolutely. That is about all I can do at the moment. If you require further help it would be best if you show some work and explain exactly where you are having difficulty. Moved to Real Analysis and Topology.
 May 22nd, 2012, 12:10 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: In need of help disk, series test, taylor, and power se You refer, in your title, to "disk" (I assume you mean "disk of convergence"_, Taylor, and power series but those only apply to series of powers of x, not to numerical series that you give here.

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