# N3 n 1 3 1 4 n3 n4 n3 consequently an xn solution

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Unformatted text preview: e convergence at endpoints But then Example 1: ﬁnd the radius of convergence of the power series ∞ n=0 4n an = n+1 an+1 4 n (−4) √ xn . n+3 (n + 1) + 3 1 √ = 4 n+3 n+4 . n+3 Consequently, ∞ an xn Solution: this is a power series with n=0 R = lim n→∞ an 1 = lim n→∞ 4 an+1 n+4 1 =. n+3 4 (−4)n 4n an = √ = (−1)n √ n+3 n+3 ∞ an xn be a general To begin to understand what the Radius of convergence does for us, let power series. Whenever the series converges it deﬁnes a function n=0 ∞ an xn . f (x) = a0 + a1 x + a2 x2 + a3 x3 + . . . + an xn + . . . = n=0 It’s certainly deﬁned at x = 0, for instance, since f (0) = a0 . Deﬁnition: the Interval of Convergence of a power series n an xn is the largest interval on which the series is deﬁned; it is the domain of th...
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## This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas at Austin.

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