Question

6. Give an example of a non-constant sequence that satisfies the given conditions or explain why such a sequence does not exist: (1) {an} is bounded above but not convergent. (2) {an} is neither decreasing nor increasing but still converges. (3) {an} is bounded but divergent. (4) {an} is unbounded but convergent. (5) {an} is increasing and converges to 2.

Answer 1

6 (1.) Let, An = (-1)", HNEN. See that, the sequence {an} = {(1) h} in [ Since, -15 ans 1, VNEN] we bounded. above Now, lim Azn lim 1 = nya & lim azn hy lim (1)=-1 ma & limagni Since, limaan {ans is not convergent. il 3-11n) is bounded above but convergent not 2 an =(-1ght! I h , then ... the sequence { an} = {(141/13 is neither decreasing nor increasing but still converges to o t lim aan =o= lim aani 204] nya no 3 Let an an = Sinhte Sinht , new the sequence ? am] = { sin heme) - {1,0,-4,0,1,0, is bounded, but not convergent. ie divergent ༣

we know that, A convergent sequence is bounced This implies that, An unboundea sequence is not Convergent. So, there exists De sequence pang which is unbounded but Convergent. Les. an=2-1 / NEIN. 1 Now, anal = 20 ntl Now, 1 & NEIN. ntl n yls - 1 then nt! a 2-5 22 1 nal an < anti VEIN. {ans ={2-4} is increasing, limana lim (2-1 / ) = 2 ie lang Converges to 2.