Question

Determine whether the following series converges absolutely, converges conditionally, or diverges. 00 (-1)+1e 3k Σ-11: -Σ ak (k 17 k 1 k 1 Find lim a. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. koo O A. lim ak koo O B. The Ilimit does not exist. (1)* 1 (k 17) 3k e Σ. Now, let denote What can be concluded from this result using the Divergence Test? k 1 O A. The series La must converge. O B. The series 2 a must diverge. O C. The series a must converge. O D. The series a must diverge. O E. The Divergence Test is inconclusive. Are the terms of the sequence a decreasing after some point? ves

Answer 1

Consider lim ak ktoo cele 200 noi (3.23* in esponential en function which has a rapid growth) (as compare to (x-17) does not exist. The limit Scanned with CamScanner

Wow, let EQka Ž 4-5* c* - KET (K-17), KH 3K The series Eak must (0) diverge Wow, lakl= (-DRt ek t ek 1 (k-17). / (K-17) This is an increasing sequence. 8. The term of sequence lak1 does not decrease. Answer is NO. Note divergence Test: If lim ax #0 , then z ak is divergent CS Scanned with CamScanner

a Kto Alternating Series Test: Consider I (1) " q ax is a decreasing sequence and lim ax = 0 , then I tt ak is convergent : Hotel Ž 1]**? 23 ,- The alternating K=1 (k-17)] series test does NOT apply conver Now Consider Elak! 3k • het om ::. \ax) in NOT #0 convergent. ...D r that is NO, Because of Divergence Flcs Scanned with CamScanner Test)

Now; {ax diverges because lim Ok &0 Kto :: or cs Scanned with CamScanner