5 k parenleftbig 2 x 6 parenrightbig k 1 interval

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Unformatted text preview: 5 k parenleftBig 2 x + 6 parenrightBig k . 1. interval conv. = parenleftBig 1 2 , 11 2 parenrightBig 2. series converges only at x = − 3 cor- rect 3. interval conv. = ( −∞ , ∞ ) 4. interval conv. = parenleftBig − 1 2 , 1 2 parenrightBig 5. interval conv. = parenleftBig − 11 2 , − 1 2 parenrightBig Explanation: rubio (mdr399) – HW14 – Radin – (57410) 7 There are three possibilities for the interval of convergence of the infinite series ∞ summationdisplay k = 0 a k ( x − a ) k . First set L = lim k →∞ vextendsingle vextendsingle vextendsingle a k +1 a k vextendsingle vextendsingle vextendsingle . Then (i) when L > 0, the interval of convergence is given by parenleftBig a − 1 L , a + 1 L parenrightBig ; (ii) when L = 0, the interval of convergence is given by ( −∞ , ∞ ); (iii) when L = ∞ , the interval of conver- gence reduces to the point { a } , i.e. , the series converges only at x = a . For the given series, ∞ summationdisplay k =0 ( − 1) k ( k + 1)! 5 k parenleftBig 2 x + 6 parenrightBig k = ∞ summationdisplay k =0 ( − 1) k ( k + 1)!2 k 5 k parenleftBig x + 3 parenrightBig k , so the corresponding value of a k is a k = ( − 1) k 2 k ( k + 1)! 5 k . In this case vextendsingle vextendsingle vextendsingle a k +1 a k vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle − ( k + 2)!2 k +1 5 k +1 slashBig ( k + 1)!2 k 5 k vextendsingle vextendsingle vextendsingle = 2( k + 2) 5 . Thus L = lim k →∞ 2( k + 2) 5 = ∞ . Consequently, series converges only at x = − 3 . keywords: 013 10.0 points If the series ∞ summationdisplay n =0 c n x n converges when x = − 3 and diverges when x = 4, which of the following series converge without further restrictions on { c n } ? A. ∞ summationdisplay n = 0 c n ( − 2) n B. ∞ summationdisplay n = 0 c n C. ∞ summationdisplay n = 0 c n ( − 3) n +1 1. none of them 2. A and B only 3. C only 4. A and C only 5. B and C only 6. A only 7. all of them correct 8. B only Explanation: A. The interval of convergence of series ∞ summationdisplay n = 0 c n x n rubio (mdr399) – HW14 – Radin – (57410) 8 contains ( − 3 , 3). Since x = − 2 belongs to this interval, the series summationdisplay n = 0 c n ( − 2) n converges also....
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