Ii determine the interval of convergence of the

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(ii) Determine the interval of convergence 3. [ 1 / 6 , 1 / 6] 4. ( −∞ , ) 5. [ 1 , 1] 6. [ 6 , 6) 7. converges only at x = 0 8. [ 6 , 6] correct Explanation: Since R = 6, the given series (i) converges when | x | < 6, and (ii) diverges when | x | > 6. On the other hand, at the point x = 6 and x = 6, the series reduces to summationdisplay n =1 ( 1) n n 4 , summationdisplay n =1 1 n 4 respectively. But summationdisplay n =1 vextendsingle vextendsingle vextendsingle ( 1) n n 4 vextendsingle vextendsingle vextendsingle = summationdisplay n =1 1 n 4 , so by the p -Series Test with p = 4 both series converge. Consequently, the given series has interval convergence = [ 6 , 6] . 010 10.0points Find the radius of convergence, R , of the power series summationdisplay n =1 n ( x 5) n .
of the series.
garcia (pgg378) – HW14 – he – (53725) 6 3. R = 4. R = 5 5. R = 0 Explanation: The given series has the form summationdisplay n =1 a n ( x 5) n where a n = n. Now for this series, (i) R = 0 if it converges only at x = 5, (ii) R = if it converges for all x , while 0 <R< (iii) if it converges for | x 5 | <R , and (iv) diverges for | x 5 | >R . But vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = n + 1 n = radicalbigg n + 1 n , so lim n → ∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = 1 . By the Ratio test, therefore, the given series (a) converges for all | x 5 | < 1, and (b) diverges for all | x 5 | > 1. Consequently, R = 1 . 011 10.0points Determine the interval of convergence of the series summationdisplay n =1 n 3 ( x 5) n . 1. interval convergence = ( 6 , 4) 2. converges only at x = 5 3. interval convergence = (4 , 6) correct 4. interval convergence = [4 , 6) 5. interval convergence = ( −∞ , ) 6. interval convergence = ( 6 , 4] Explanation: The given series has the form summationdisplay n =1 a n ( x 5) n with a n = n 3 . Now lim n →∞ a n +1 a n = lim n →∞ parenleftBig n + 1 n parenrightBig 3 = 1 . By the Ratio Test, therefore, the given series (i) converges when | x 5 | < 1, and (ii) diverges when | x 5 | > 1. On the other hand, at the points x 5 = 1 and x 5 = 1 the series reduces to summationdisplay n =1 ( 1) n n 3 , summationdisplay n =1 n 3 respectively. But by the Divergence Test, both of these diverge. Consequently, interval convergence = (4 , 6) . 012 10.0points Find the radius of convergence, R , of the series summationdisplay n =1 ( 1) n n 3 n ( x + 1) n . 1. R = 1 3
garcia (pgg378) – HW14 – he – (53725) 7 2. R = 0 3. R = 3 correct 4. R = 5. R = 1 Explanation: The given series has the form summationdisplay n =1 a n ( x + 1) n with a n = ( 1) n n 3 n . Now for this series (i) R = 0 if it converges only at x = 0, (ii) R = if it converges for all x , while if R> 0, (iii) it coverges when | x + 1 | <R , and (iv) diverges when | x + 1 | >R . But lim n → ∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n → ∞ n 3( n + 1) = 1 3 . By the Ratio Test, therefore, the given series converges when | x +1 | < 3 and diverges when | x + 1 | > 3. Consequently, R = 3 . 013 10.0points Determine the radius of convergence, R , of the power series summationdisplay n =1 ( 2) n n ( x + 5) n .

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