# C true if summationdisplay a n were convergent then

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Unformatted text preview: C. True: if summationdisplay | a n | were convergent, then summationdisplay a n would be absolutely convergent, hence convergent. 022 10.0 points If the radius of convergence of the power series ∞ summationdisplay n =0 c n x n is 9, what is the radius of convergence, R , of the power series ∞ summationdisplay n = 0 c n x 2 n ? 1. R = ∞ 2. R = 0 3. R = 3 correct 4. R = 9 5. R = 1 6. R = √ 3 Explanation: As the radius of convergence of the series (1) ∞ summationdisplay n =0 c n x n is 9, this means that series (1) (i) converges when | x | < 9, and (ii) diverges when | x | > 9. But then the series (2) ∞ summationdisplay n = 0 c n x 2 n will (i) converge when | x 2 | < 9, and (ii) diverges when | x 2 | > 9. Consequently, series (2) (i) converges when | x | < 3, and (ii) diverges when | x | > 3, and so series (2) will have radius of conver- gence R = 3 . keywords: PowerSeries, PowerSeriesExam, 023 10.0 points If the series ∞ summationdisplay n =0 c n x n converges when x =- 4 and diverges when x = 6, which of the following series must converge without further restrictions on { c n } ? A. ∞ summationdisplay n = 0 c n (- 4) n +1 B. ∞ summationdisplay n = 0 (- 1) n c n 7 n 1. neither of them 2. both of them 3. B only 4. A only correct Explanation: A. Since ∞ summationdisplay n = 0 c n (- 4) n +1 =- 4 parenleftBigg ∞ summationdisplay n =0 c n (- 4) n parenrightBigg cruz (fmc326) – HW03 – kalahurka – (55295) 12 the series ∞ summationdisplay n =0 c n (- 4) n +1 converges. B. The series ∞ summationdisplay n =0 c n x n diverges for all | x | > 6, hence for x =- 7. 024 10.0 points Determine the interval of convergence of the power series ∞ summationdisplay n = 1 (- 2) n 3 √ n ( x- 5) n . 1. interval of cgce = parenleftBig- 11 2 ,- 9 2 parenrightBig 2. interval of cgce = parenleftBig- 11 2 ,- 9 2 bracketrightBig 3. interval of cgce = bracketleftBig 9 2 , 11 2 parenrightBig 4. interval of cgce = bracketleftBig- 11 2 ,- 9 2 bracketrightBig 5. interval of cgce = parenleftBig 9 2 , 11 2 bracketrightBig correct 6. interval of cgce = bracketleftBig- 11 2 ,- 9 2 parenrightBig 7. interval of cgce = bracketleftBig 9 2 , 11 2 bracketrightBig 8. interval of cgce = parenleftBig 9 2 , 11 2 parenrightBig Explanation: The given series has the form ∞ summationdisplay n =1 (- 1) n a n ( x- 5) n where a n = 2 n 3 √ n . But then vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 2 parenleftBig 3 √ n 3 √ n + 1 parenrightBig = 2 parenleftBig 3 radicalbigg n n + 1 parenrightBig . in which case, lim n →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 2 . By the Ratio test, therefore, the given series (i) converges when | x- 5 | < 1 / 2, and (ii) diverges when | x- 5 | > 1 / 2; in particular, it converges when- 1 2 + 5 < x < 1 2 + 5 , i.e. , on the interval parenleftbigg 9 2 , 11 2 parenrightbigg . To check for convergence at the endpoints of this interval, observe first that when x = 9 2 , the series becomes ∞ summationdisplay n =1 1 3 √ n which diverges by the p-series test....
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