1 summationdisplay n 0 c n 8 n is divergent 2 summationdisplay n 0 c n 6 n is

1 summationdisplay n 0 c n 8 n is divergent 2

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? 1. summationdisplay n =0 c n ( - 8) n is divergent 2. summationdisplay n =0 c n ( - 6) n is convergent 3. summationdisplay n =0 c n ( - 8) n is convergent 4. summationdisplay n =0 c n ( - 6) n is divergent 014 10.0 points Find the interval of convergence of the power series summationdisplay n =1 2 x n n . 1. interval of cgce = [ - 2 , 2) 2. interval of cgce = ( - 2 , 2) 3. interval of cgce = [ - 2 , , 2] 4. interval of cgce = [ - 1 , 1] 5. interval of cgce = ( - 1 , 1) 6. interval of cgce = ( - 1 , 1] 7. interval of cgce = ( - 2 , 2] 8. interval of cgce = [ - 1 , 1) 015 10.0 points Determine the interval of convergence of the series summationdisplay n =1 ( - 1) n 2 n ( n + 3) x n . 1. converges only at x = 0 2. none of the other answers 3. interval convergence = ( - 2 , 2] 4. interval convergence = parenleftBig - 1 2 , 1 2 bracketrightBig 5. interval convergence = bracketleftBig - 1 3 , 1 3 parenrightBig 6. interval convergence = ( -∞ , ) 7. interval convergence = [ - 3 , 3) 016 (part 1 of 2) 10.0 points For the series summationdisplay n =1 x n ( n + 4)! , (i) determine its radius of convergence, R . 1. R = 2. R = 0 3. R = 1 4. R = 4 5. R = 1 4 017 (part 2 of 2) 10.0 points
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husain (aih243) – HW 3 – mann – (54675) 4 (ii) Determine the interval of convergence of the series. 1. interval convergence = ( - 4 , 4) 2. interval convergence = [ - 4 , 4] 3. interval convergence = [ - 1 , 1] 4. converges only at x = 0 5. interval convergence = ( - 1 , 1] 6. interval convergence = ( -∞ , ) 018 10.0 points Find the interval of convergence of the power series summationdisplay n =0 5 n 3 n 5 + 7 x n .
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  • Fall '07
  • Sadler
  • Mathematical Series, Radius of convergence, interval convergence, 1 3k

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