X 25 and ii diverges when x 25 but then the series 2

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Unformatted text preview: x | < 25, and (ii) diverges when | x | > 25. But then the series (2) ∞ summationdisplay n = 0 c n x 2 n will (i) converge when | x 2 | < 25, and (ii) diverges when | x 2 | > 25. Consequently, series (2) (i) converges when | x | < 5, and (ii) diverges when | x | > 5, and so series (2) will have radius of conver- gence R = 5 . keywords: PowerSeries, PowerSeriesExam, 023 10.0 points If the series ∞ summationdisplay n =0 c n x n converges when x =- 3 and diverges when x = 5, which of the following series must converge without further restrictions on { c n } ? A. ∞ summationdisplay n = 0 c n (- 2) n B. ∞ summationdisplay n = 0 c n (- 3) n +1 1. neither of them 2. A only 3. B only 4. both of them correct Explanation: A. The interval of convergence of series ∞ summationdisplay n =0 c n x n contains (- 3 , 3). Since x =- 2 belongs to this interval, the series summationdisplay n =0 c n (- 2) n converges also. B. Since ∞ summationdisplay n = 0 c n (- 3) n +1 =- 3 parenleftBigg ∞ summationdisplay n =0 c n (- 3) n parenrightBigg the series ∞ summationdisplay n =0 c n (- 3) n +1 converges. 024 10.0 points Determine the interval of convergence of the power series ∞ summationdisplay n = 1 (- 3) n √ n ( x- 1) n . keller (mjk2535) – HW03 – kalahurka – (55250) 12 1. interval of cgce = bracketleftBig 2 3 , 4 3 bracketrightBig 2. interval of cgce = bracketleftBig- 4 3 ,- 2 3 bracketrightBig 3. interval of cgce = bracketleftBig- 4 3 ,- 2 3 parenrightBig 4. interval of cgce = parenleftBig 2 3 , 4 3 parenrightBig 5. interval of cgce = parenleftBig- 4 3 ,- 2 3 parenrightBig 6. interval of cgce = parenleftBig 2 3 , 4 3 bracketrightBig correct 7. interval of cgce = parenleftBig- 4 3 ,- 2 3 bracketrightBig 8. interval of cgce = bracketleftBig 2 3 , 4 3 parenrightBig Explanation: The given series has the form ∞ summationdisplay n =1 (- 1) n a n ( x- 1) n where a n = 3 n √ n . But then vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 3 parenleftBig √ n √ n + 1 parenrightBig = 3 parenleftBig radicalbigg n n + 1 parenrightBig . in which case, lim n →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 3 . By the Ratio test, therefore, the given series (i) converges when | x- 1 | < 1 / 3, and (ii) diverges when | x- 1 | > 1 / 3; in particular, it converges when- 1 3 + 1 < x < 1 3 + 1 , i.e. , on the interval parenleftbigg 2 3 , 4 3 parenrightbigg . To check for convergence at the endpoints of this interval, observe first that when x = 2 3 , the series becomes ∞ summationdisplay n =1 1 √ n which diverges by the p-series test. On the other hand, when x = 4 3 , the series becomes ∞ summationdisplay n =1 (- 1) n √ n which converges by the Alternating Series Test. Consequently, the given series has interval of cgce = parenleftBig 2 3 , 4 3 bracketrightBig ....
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