{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M408L_HW13

# M408L_HW13 - hyun(hh7953 HW13 gogolev(57440 This print-out...

This preview shows pages 1–3. Sign up to view the full content.

hyun (hh7953) – HW13 – gogolev – (57440) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine whether the series 2 3 2 4 + 2 5 2 6 + 2 7 . . . is conditionally convergent, absolutely con- vergent or divergent. 1. series is conditionally convergent cor- rect 2. series is divergent 3. series is absolutely convergent Explanation: In summation notation, 2 3 2 4 + 2 5 2 6 + 2 7 . . . = summationdisplay n =1 ( 1) n 1 f ( n ) , with f ( x ) = 2 x + 2 . Now the improper integral integraldisplay 1 f ( x ) dx = integraldisplay 1 2 x + 2 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 2 n + 2 > 2 n + 1 + 2 = f ( n + 1) , while lim n →∞ 2 n + 2 = 0 . Consequently, by the Alternating Series Test, the given series is conditionally convergent . keywords: alternating series, Alternating se- ries test, conditionally convergent, absolutely convergent, divergent 002 10.0 points Which one of the following series is conver- gent? 1. summationdisplay n =1 5 6 + n 2. summationdisplay n =1 ( 1) n 1 4 + n correct 3. summationdisplay n =1 ( 1) 2 n 6 5 + n 4. summationdisplay n =1 ( 1) 3 6 5 + n 5. summationdisplay n =1 ( 1) n 1 4 + n 6 + n Explanation: Since summationdisplay n =1 ( 1) 3 6 5 + n = summationdisplay n =1 6 5 + n , use of the Limit Comparison and p -series Tests with p = 1 2 shows that this series is divergent. Similarly, since summationdisplay n =1 ( 1) 2 n 6 5 + n = summationdisplay n =1 6 5 + n , the same argument shows that this series as well as summationdisplay n =1 5 6 + n is divergent. On the other hand, by the Divergence Test, the series summationdisplay n =1 ( 1) n 1 4 + n 6 + n

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
hyun (hh7953) – HW13 – gogolev – (57440) 2 is divergent because lim n → ∞ ( 1) n 1 4 + n 6 + n negationslash = 0 . This leaves only the series summationdisplay n =1 ( 1) n 1 4 + n . To see that this series is convergent, set b n = 1 4 + n . Then (i) b n +1 b n , (ii) lim n → ∞ b n = 0 . Consequently, by the Alternating Series Test, the series summationdisplay n =1 ( 1) n 1 4 + n is convergent. 003 10.0 points Determine whether the series summationdisplay n =1 ( 1) n 1 e 1 /n 4 n is absolutely convergent, conditionally con- vergent or divergent. 1. conditionally convergent correct 2. absolutely convergent 3. divergent Explanation: Since summationdisplay n =1 ( 1) n 1 e 1 /n 4 n = 1 4 summationdisplay n =1 ( 1) n e 1 /n n , we have to decide if the series summationdisplay n =1 ( 1) n e 1 /n n is absolutely convergent, conditionally con- vergent, or divergent. First we check for absolute convergence.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

M408L_HW13 - hyun(hh7953 HW13 gogolev(57440 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online