Question

4. [-/10 Points] DETAILS SCALCET8 11.8.AE.002. MY NOTES EXAMPLE 2 For what values of x does the series § (x – 3)" converge? n = 1 SOLUTION Let a, = (x - 3)"/n. Then an + 1 n (x - 3)" 1x - 3 → as n 00. and divergent when x - 3 > Now By the Ratio test, the given series is absolutely convergent, and therefore convergent, when |x - 3|< |x - 3| <1 | <x - 3< 1<x</ so the series converges when and diverges when x< or x> The Ratio Test gives no information when X – 31 = 1, so we must consider x = 2 and x = 4 separately. If we put x = 4 in the series, it becomes E 1/n, the harmonic series, which is divergent. If x = 2, the series is 3 (-1)"/n, which --Select--- by the Alternating Series Test. Thus the given power series converges for 2?x? 4. 5. [0/10 Points] DETAILS PREVIOUS ANSWERS SCALCET8 11.8.516.XP. MY NOTES Find the radius of convergence, R, of the series. n(x - 7)" n3 + 1 n = 1 R = | + -14 calcPad Operations Functions Symbols Find the interval, I, of convergence of the series. (Enter your answer using interval notation.) Relations nd I = Sets x VO ! Vectors Submit Answer Trig Groot

Answer 1

Ex. 2; E (2-3) 1 C-3 Som! let an= anti (x-3)+1 n - ab (5+1) 632-35 T 12-31 o suego - -12-311 (1+) 1 X-3131 > T- < 2-3 = 1 2 < s <4 so the series converge alben 2<x<4 and diverge lahen x < 2 or x>4. E Which is convergent ect-zoda soll 2 MS ele Ttale iselos let ana GCE-ze) +12(2-7) ity & Ttak 32+03+1 well anti lim با 0 tast-so) CHO rte chied Ptd) cho urte) e It-sel

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