4. Determine if the series converges absolutely, converges conditionally, or diverges. (Show all of your work!) (a) 22:2 2:1?\" 00 1 \"'1 (b) 271:1 ( % (71)\"(n'i'lllt4 (Cl 220:1 "2 Ratio Test suppose we have the Series Zan, Define L = lim anti an Then if LCI, the series is absolutely convergent if LDI . the series is divergent if 1= 1 , the series is divergent, conditionally convergent (or) absolutely convergent Proof, let there be some Numbers f Such that and assume L lantil 21aul lantk / lantk- 1 1 * laul Now we have, lantil lanl where K = 0, 1 , 2 .-.00 2 lanlak ) This is a geometric series Al also, 0 N Do Elanl = lault & lan! nel n = 1 MENtl . finite E laul is convergent n = 1 oure assumption was correct LEI, Series converges.4 S Glin 1= 2 here summation starks facem n= 2 and goes to Do " we will change the limits from n= 1 to do . : Power with term will be I less . DO ( - 1) 1 - 1 an = 1 net ( 71 - 1 ) en ( u - 1 ) Ratio Test anti = n = 1 n lun L = lim (-12" ( U-1)en ( n- ) / n - (n - 1 ) ( - 1) &n Ch-1 ) / ": n is very large = lin 1 - n Rim * ( 1 - 1 ) W 12-1 1 L = series can be convergent, divergent.b M n = 1 Ju Ratio Test autie Con Ju L = Jim m - Unti Gin - L = Jim Unti dim - Vu Jn ( Itt ) 2 dim JX L = 1 Series may be convergent on divergentn ( nti) / 4 C 1=1 1 2 Ratio Test ( on + 1) (n+ 2)/ 4 . oz Jim (-1) L= 170 dim ( - 1 ) ntly ( 1 + 2 - " ) n ) 2 ( in + 1 ) 2 lim ( - 1) ( 4 ( FREI ) ? lim lim Not Bracket it is a modulus operator = lim Um L = 1 Series may be convergent or divergent