1.)

Use the Limit Comparison Test to determine whether the series converges or diverges. 31:1 =1 00 00 1 The comparison series is Z 5 = 2 6(7) where c = and p 2 71:1 71:1 T1 Then, ljm n 2 713430 1'}. 00 00 2 5,1 is a ? v , therefore 2 an ? v by the Limit 5'1: n=1 Comparison Test. Use the Limit Comparison Test to determine whether the series converges or diverges. n=2 71:2 TI. 00 00 1 The comparison series is Z bu = 2 ((p) where c = and 'p 2 11:2 1122 R Then, Hm n = 1'1}DO n 00 00- 2 bin is a ? v , therefore 2 an ? v by the Limit 14:2 11:2 Comparison Test. Use the Limit Comparison Test to determine whether the series converges or diverges. oo oo 9'18\" 00 00 The comparison series is Z 5 = Z cairn1 where a = and \"r 2 11:1 11:1 :1 Then, Hm n = nmo bu 0C} 00 2 an is a ? v , therefore Zen ? v by the \"=1 n=1 Limit Comparison Test. Use the Limit Comparison Test to determine whether the series converges or diverges. 00 0 v9n2+2n+9 2a: :1 12?15 i 271,5 i 6n OO 00 1 The comparison series is Z on = Z c (33) where c = and p 2 71:1 71:1 TI. Then. liJIla = \"rtmo b: 00- 00 Z 6,. is a '? v ,therefore Zen ? v by the Limit n=1 Comparison Test. Use the Limit Comparison Test to determine whether the series converges or diverges. f: :0: 3?;3 i 971 1 on = n=1 n=l {TL + 2)6 00 00 1 The comparison series is Z b = 2 ((p) where c = and 'p 2 n=1 n=1 \"R Then, Hm n = 1'1}DO 1'1. 00 00 Z on is a ? v , therefore Z on ? v by the Limit =1 n=1 Comparison Test. Use the Limit Comparison Test to determine whether the series converges or diverges. OO 8 . 11" + 8 . 2n an 13" + 6n n=1 n= The comparison series is bn = aph-1 where a = and r = n=1 n=1 an Then, lim bn E bn is a ? v, therefore an v by the n=1 n=] Limit Comparison Test.Use the Limit Comparison Test to determine whether the series converges or diverges. DO 4n + 11 an n=1 n=1 V18nll + 3n3 OO OO The comparison series is Ebn = Ec( me) where c = and p n=1 n=1 Then, lim an bn is a ? v , therefore an by the Limit n=1 ? n=1 Comparison convergent p-series divergent p-seriesUse the Limit Comparison Test to determine whether the series converges or diverges. 00 00 7 gar}: 7n+38\" 00 00 The comparison series is Z I) = Zorn1 where a = and 'r 2 n=1 n=1 G Then, Em A = nmo bu 00 00 25,; is a ? v , therefore 2 an ? v by the '11:]. 11:1 Limit Comp . . convergent geometric series divergent geometric series Use the Limit Comparison Test to determine whether the series converges or diverges. DO Can gen + 5 nen + 2 n=1 n=1 The comparison series is M bn= where c = and p = n=1 Then, lim an by is a ? v, therefore M an ? by the Limit n=1 ? n=1 Comparison convergent p-series divergent p-seriesSuppose Z on and Z 5.. are series with positive terms and 2 b... is known to be divergent. If an > on for all n, what can you say about :3 an? {3 A. Z on diverges by the Direct Comparison Test. C) B. We cannot say anything about 2 an. {3 C. E on converges by the Direct Comparison Test. Suppose :3 an and Z 5,. are series with positive terms and :3 b... is known to be convergent. If an 2 b... for all n, what can you say about 2 on? {3 A. E on diverges by the Direct Comparison Test. C) B. E on converges by the Direct Comparison Test. (3 C. We cannot say anything about :3 an. Suppose Z on and Z 5.. are series with positive terms and 2 b... is known to be divergent. If on c ( mix ) wi where c = and p = n=1 , which means an ? v bn for all n 2 1. bn is a ? v therefore an ? by the Direct n=1 ? n=1 Comparison convergent p-series divergent p-seriesUse the Direct Comparison Test to determine whether the series converges or diverges. 0 _ 0 titan1 n 2% Z 32n n=3 n=3 00 00 The comparison series is E b = E urn1 where a = and a" 2 n=3 n=3 which means an ? v I)\" for all n 2 3. 00 00 :5\" is a '? v , therefore Zen ? v by the \"=3 n=3 Direct Com convergent geometric series divergent geometric series Use the Alternating Series Test to determine whether the series converges or diverges. (For limits, enter a number, "-infnity , Infinity", or "DNE" as appropriate.) 0 A. {an} is decreasing because it has a constant numerator and increasing denominator, therefore the series converges by the Alternating Series test. O B. 11111 on = , so the series diverges by the Divergence Test. nPOO Use the Alternating Series Test to determine whether the series converges or diverges. (For limits, enter a number, "-infnity , Infinity", or "DNE" as appropriate.) f.(1>\"('\"ff\") n lim 5,. = 11}00 O A. {bu} is ultimately decreasing because the function f satifying n) = 5,1 is decreasing on the interval Therefore the series converges by the Alternating Series test. 0 B. 1111] on = , so the series diverges by the Divergence Test. 71}00 Use the Alternating Series Test to determine whether the series converges or diverges. (For limits, enter a number, "-infnityr , Infinity", or "DNE" as appropriate.) 0 A. {on} is ultimately decreasing because the function f satifying n) = b\" is decreasing on the interval Therefore the series converges by the Alternating Series test. 0 B. 1111] an = , so the series diverges by the Divergence Test. 11,}00