{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Gilbert_Hmwk05sol

# Gilbert_Hmwk05sol - moseley(cmm3869 – HW05 – Gilbert...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: moseley (cmm3869) – HW05 – Gilbert – (56380) 1 This print-out should have 21 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 3 2 − 3 3 + 3 4 − 3 5 + 3 6 − . . . is conditionally convergent, absolutely con- vergent or divergent. 1. series is conditionally convergent cor- rect 2. series is absolutely convergent 3. series is divergent Explanation: In summation notation, 3 2 − 3 3 + 3 4 − 3 5 + 3 6 − . . . = ∞ summationdisplay n =1 ( − 1) n − 1 f ( n ) , with f ( x ) = 3 x + 1 . Now the improper integral integraldisplay ∞ 1 f ( x ) dx = integraldisplay ∞ 1 3 x + 1 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 3 n + 1 > 3 n + 1 + 1 = f ( n + 1) , while lim n →∞ 3 n + 1 = 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 ( − 1) 3 4 3 + √ n 2. ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n 4 + √ n 3. ∞ summationdisplay n =1 3 4 + √ n 4. ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n correct 5. ∞ summationdisplay n =1 ( − 1) 2 n 4 3 + √ n Explanation: Since ∞ summationdisplay n =1 ( − 1) 3 4 3 + √ n = − ∞ summationdisplay n =1 4 3 + √ 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 4 3 + √ n = ∞ summationdisplay n =1 4 3 + √ n , the same argument shows that this series as well as ∞ summationdisplay n =1 3 4 + √ n is divergent. On the other hand, by the Divergence Test, the series ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n 4 + √ n moseley (cmm3869) – HW05 – Gilbert – (56380) 2 is divergent because lim n →∞ ( − 1) n − 1 1 + √ n 4 + √ n negationslash = 0 . This leaves only the series ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n . To see that this series is convergent, set b n = 1 1 + √ 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 1 + √ n is convergent. 003 10.0 points Determine whether the series ∞ summationdisplay n =1 ( − 1) n − 1 e 1 /n 3 n is absolutely convergent, conditionally con- vergent or divergent. 1. absolutely convergent 2. divergent 3. conditionally convergent correct Explanation: Since ∞ summationdisplay n =1 ( − 1) n − 1 e 1 /n 3 n = − 1 3 ∞ 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

Gilbert_Hmwk05sol - moseley(cmm3869 – HW05 – Gilbert...

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

View Full Document
Ask a homework question - tutors are online