Lesson 26a - Power Series

Thefulldistanceis2sothe radiusis1 example2

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Unformatted text preview: mine the radius of convergence for the following: xn n2 n 1 Here we see no functions to functions, so we will use the ratio test. an 1 x n 1 n 2 lim lim 2 n n ( n 1) n a x n an 1 n2 x lim lim n a n ( n 1) 2 n an 1 n2 lim | x | lim n a n ( n 1) 2 n The highest power is n2, so we divide numerator and denominator by n2 to get: an 1 n a n lim n2 n2 | x | lim | x | n n 1 ( )2 nn We know that the ratio test will tell us it will surely converge when the limit is less than 1, this gives: |x| < 1. This means that ‐1 < x < 1. The full distance is 2, so the radius is 1. Example 2 Determine the radius of convergence for the following: nn xn n 1 Here we see functions to functions, so we will use the root test. lim an 1/ n n lim n x n n 1/ n n lim an lim an n 1/ n lim nx n 1/ n | x | lim n n n This limit goes to infinity, so the only way to get something less than 1 is if x = 0. (If we multiply nn by 0 it will become 0, and we will have a very uninteresting 0 series). This means that the radius of convergence is 0. Strategy: Find the Interval of Convergence Once we find the radius of convergence, then we know that the interval...
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This note was uploaded on 02/07/2014 for the course MATH 1004 taught by Professor Mark during the Summer '00 term at Carleton CA.

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