121616949-math.274 - 1/L 1/L and perhaps will extend to one...

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260 Chapter 10 Sequences and Series with the understanding that a n may depend on n but not on x . EXAMPLE 10.44 X n =1 x n n is a power series. We can investigate convergence using the ratio test: lim n →∞ | x | n +1 n + 1 n | x | n = lim n →∞ | x | n n + 1 = | x | . Thus when | x | < 1 the series converges and when | x | > 1 it diverges, leaving only two values in doubt. When x = 1 the series is the harmonic series and diverges; when x = - 1 it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of X n =1 x n n as a function from the interval [ - 1 , 1) to the real numbers. A bit of thought reveals that the ratio test applied to a power series will always have the same nice form. In general, we will compute lim n →∞ | a n +1 || x | n +1 | a n || x | n = lim n →∞ | x | | a n +1 | | a n | = | x | lim n →∞ | a n +1 | | a n | = L | x | , assuming that lim | a n +1 | / | a n | exists. Then the series converges if L | x | < 1, that is, if | x | < 1 /L , and diverges if | x | > 1 /L . Only the two values x = ± 1 /L require further investigation. Thus the series will definitely define a function on the interval (
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Unformatted text preview: 1 /L, 1 /L ), and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if L = 0 the limit is 0 no matter what value x takes, so the series converges for all x and the function is defined for all real numbers. If L = ∞ , then no matter what value x takes the limit is infinite and the series converges only when x = 0. The value 1 /L is called the radius of convergence of the series, and the interval on which the series converges is the interval of convergence . Consider again the geometric series, ∞ X n =0 x n = 1 1-x . Whatever benefits there might be in using the series form of this function are only avail-able to us when x is between-1 and 1. Frequently we can address this shortcoming by modifying the power series slightly. Consider this series: ∞ X n =0 ( x + 2) n 3 n = ∞ X n =0 ± x + 2 3 ¶ n = 1 1-x +2 3 = 3 1-x ,...
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