Then summationdisplay n a n summationdisplay n 1 n so by the p series test with

# Then summationdisplay n a n summationdisplay n 1 n so

This preview shows page 2 - 5 out of 12 pages. Subscribe to view the full document. Subscribe to view the full document.

#### You've reached the end of your free preview.

Want to read all 12 pages?

Unformatted text preview: Then summationdisplay n | a n | = summationdisplay n 1 n , so by the p-series test with p = 1, the series summationdisplay | a n | diverges. On the other hand, summationdisplay n a n = summationdisplay n (- 1) n n converges by the Alternating Series Test. (B) TRUE: to say that ∑ n a n converges is to say that the limit lim n →∞ S n of its partial sums S n = a 1 + a 2 + . . . + a n converges. But then lim n →∞ a n = lim n →∞ ( S n- S n − 1 ) = 0 . 004 10.0 points Determine which, if either, of the series A. ∞ summationdisplay n =1 1 n ! cos parenleftBig n π 3 parenrightBig B. ∞ summationdisplay n =1 (- 1) n − 1 5 n 2 n 2 + 3 are absolutely convergent. 1. neither of them 2. both of them 3. A only correct 4. B only Explanation: A. Absolutely convergent: use Comparison Test with a n = 1 n ! vextendsingle vextendsingle vextendsingle cos parenleftBig n π 3 parenrightBigvextendsingle vextendsingle vextendsingle , b n = 1 n ! . cadena (jc59484) – HW13 – lawn – (55930) 3 For then < a n ≤ b n since | cos( x ) | ≤ 1 for all x . But lim n →∞ b n +1 b n = lim n →∞ 1 n + 1 = 0 < 1 . So by the Ratio Test the series ∞ summationdisplay n =1 1 n ! converges, and hence the given series is absolutely convergent. B. Not absolutely convergent: apply the Limit Comparison Test with a n = 5 n 2 n 2 + 3 , b n = 1 n . But by the p-series Test with p = 1, the series summationdisplay b n diverges. 005 10.0 points Determine which, if any, of the series A. ∞ summationdisplay m =1 ( m √ 6- 1) m B. ∞ summationdisplay m =1 4 e − 5 m m ! are convergent. 1. both of them 2. B only 3. A only correct 4. neither of them Explanation: A. The given series can be written in the form ∞ summationdisplay m =1 a m where a m = ( m √ 6- 1) m . But then m radicalbig | a m | = 6 1 /m- 1 , in which case lim m →∞ m radicalbig | a m | = 1- 1 = 0 . Consequently, by the Root Test, the given series converges . B. The given series has the form ∞ summationdisplay m =1 a m where a m = 4( m !) e 5 m . But then vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = ( m + 1)! m ! e 5 m e 5 m +5 = m + 1 e 5 . Thus lim m →∞ vextendsingle vextendsingle vextendsingle vextendsingle a m +1 a m vextendsingle vextendsingle vextendsingle vextendsingle = ∞ . Consequently, by the Ratio Test, the given series is divergent . 006 10.0 points Decide whether the series ∞ summationdisplay k =3 k ln k ( k + 5) 3 is convergent or divergent. 1. divergent 2. convergent correct cadena (jc59484) – HW13 – lawn – (55930) 4 Explanation: Since < k ln k ( k + 5) 3 < k ln k k 3 = ln k k 2 , for k ≥ 3, the Comparison Test ensures that the given series converges if the series ∞ summationdisplay k =3 ln k k 2 converges. But by the Integral test, this last series converges if the improper integral integraldisplay ∞ 3 ln x x 2 dx converges....
View Full Document

• Fall '11
• Gramlich
• Accounting, Mathematical Series, lim
• • •  