263c Math Worksheet 8.5 Power Series Due: Monday April 16th Work on these problems in groups of 2-3 and turn them in together.
1. Find the radius of convergence and interval of convergence of the series. (a) (b) (c) (d) (e) (f) 2. If xn n7n xn n10 10n n!(2x - 1)n (-1)n (-1)n x2n (2n)! xn 2n ln n
n(x - 2)n n3 - 1
cn 9n is convergent, does it follow that
n=0
cn (-3)n is convergent?
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Solutions 1. In all cases, we will determine the radius of convergence by using the Ratio Test. an+1 n+1 1 = |x| |x| R = 7. (a) an 7n 7 Since f is centered about zero, the interval of convergence is at least (-7, 7). Now we test the endpoints of the interval. (-1)n . When x = -7; f (-7) = n This is the alternating harmonic series, which is convergent. 1 When x = 7; f (7) = . n This is the harmonic series, which is divergent. Hence, the interval of convergence is IOC=[-7, 7). an+1 n10 1 = R = 10. |x| |x| 10 an 10(n + 1) 10 Since f is centered about zero, the interval of convergence is at least (-10, 10). Now we test the endpoints of the interval. (-1)n . When x = -10; f (-10) = n10 This is convergent by the Alternating Series Test. (Alternatively, |an | is a p-series (b) an is absolutely convergent, and hence convergent). 1 When x = 10; f (10) = . n10 This is a p-series with p = 10, which is convergent. Hence, the interval of convergence is IOC=[-10, 10]. (c) an+1 = (n + 1)|2x - 1| (when x = 1/2) an Hence, the interval of convergence is IOC={1/2}. R = 0. with p = 10, whence
an+1 1 = |x|2 0 R = . an (2n + 2)(2n + 1) Hence, the interval of convergence is IOC=(-, ). (d) (e) an+1 n ln 1 = |x| |x| R = 2. an 2 ln(n + 1) 2 Since f is centered about zero, the interval of convergence is at least (-2, 2). Now we test the endpoints of the interval. 1 When x = -2; f (-2) = . ln n To test convergence, we will use the Comparison Test.
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1 is divergent by the Integral Test, then according to the n ln n 1 is divergent, too. ln n (-1)n When x = 2; f (2) = . ln n This is convergent by the Alternating Series Test. Hence, the interval of convergence is IOC=(-2, 2]. an+1 (n + 1)(n3 - 1) |x - 2| |x - 2| R = 1. = an n[(n + 1)3 - 1] Since f is centered about two, the interval of convergence is at least (1, 3). Now we test the endpoints of the interval. (-1)n n . When x = 1; f (1) = n3 - 1 This is convergent by the Alternating Series Test. n When x = 3; f (3) = . 3-1 n This series is a bit tricky, so pay attention. We will test for convergence by using the Comparison Test. Since 3n2 3n we can add -3n2 + 3n to the denominator, which will make our fraction larger; (f ) n n n 3 = n3 - 1 n - 3n2 + 3n - 1 (n - 1)3 We find that 1 is convergent by using the Integral Test. (n - 1)3 Let u = n - 1. Then n dn = (n - 1)3 Whence, u+1 du = u3 1 du + u2 1 du < u3
1 1 . Since ln n n ln n Comparison Test,
n is convergent by the Comparison Test. n3 - 1 Hence, the interval of convergence is IOC=[1, 3].
2. Yes. Consider the function f (x) = cn xn . This is a power series about zero. Since f (9) is convergent, this implies that the domain of f includes the interval (-9, 9]. In particular, -3 is in the domain of f . Consequently, f (-3) = cn (-3)n must be convergent.
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