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Question Number 192738 by Mastermind last updated on 25/May/23

Find the supremum and infimum  of each of the following sequence    a) {((n−1)/(2n))}    b) {(((−)^n n)/(2n+1))}    c){((1+(−)^n )/3)}    d) {sin((nπ)/2)}    e) {(1/n) − sin((nπ)/2)}    f) {(1+(1/(2n)))cos((nπ)/3)}    Help!

Findthesupremumandinfimumofeachofthefollowingsequencea){n12n}b){()nn2n+1}c){1+()n3}d){sinnπ2}e){1nsinnπ2}f){(1+12n)cosnπ3}Help!

Answered by Rajpurohith last updated on 26/May/23

a)1     b)sup=(1/2) and inf=−(1/2)  c)sup=(2/3)  and inf=(1/3)  d)sup=1 and inf=−1  e)sup=2 and inf=0  f)sup=(3/2) and inf=1

a)1b)sup=12andinf=12c)sup=23andinf=13d)sup=1andinf=1e)sup=2andinf=0f)sup=32andinf=1

Commented by Mastermind last updated on 26/May/23

Thank you but i need solution  how you got each

Thankyoubutineedsolutionhowyougoteach

Commented by MM42 last updated on 27/May/23

sir  your answers are incorrect.please pay atyention  my answers

siryouranswersareincorrect.pleasepayatyentionmyanswers

Answered by MM42 last updated on 27/May/23

a)0,(1/4),(1/3),(3/8),...→(1/2)  ⇒inf=0  &  sup=(1/2)  b)  for the “n=odd ” −(1/3),−(3/7),−(5/(11)),...→inf=−(1/2) &  sup=−(1/3)   (i)  for the “n=even” (2/5),(4/9),(6/(13)),...→inf=(2/5)  &  sup=(1/2)   (ii)  (i) , (ii) ⇒inf=−(1/2)  &  sup=(1/2)      c)  for the “n=odd ” 0,0,0,...→inf=sup=0  for the “n=even ” (2/3),(2/3),(2/3),...→inf=sup=(2/3)  ⇒inf=0  &  sup=(2/3)  d)1,0,−1,0,1,0,−1,...⇒ inf=−1 & sup=1  e)  for the “n=even”e_n =(1/n)⇒ (1/2),(1/4),(1/6),...→0⇒inf=0  &  sup=(1/2)   (i)  for the “n=4k+1 ”e_n =(1/n)−1⇒0,− (4/5),−(8/9)...→−1⇒inf=−1  &  sup=0   (ii)  for the “n=4k+3 ”e_n =(1/n)+1⇒ (4/3),(8/7),((12)/(11))...→1⇒inf=1  &  sup=(4/3)   (iii)  (i),(ii),(iii)⇒inf=−1 & sup=(4/3)  f)  for the “n=6k ”f_n =1+(1/(2n))⇒((13)/(12)), ((25)/(24)),((37)/(36)),...→1⇒inf=1  &  sup=((13)/(12))   (i)  for the “n=6k−3 ”f_n =−(1+(1/(2n)))⇒−(7/6), −((19)/(18)),−((31)/(30)),...→−1⇒inf=−(7/6)  &  sup=−1   (ii)  for the “n=1,5,7,11,13,..”f_n =(1/2)(1+(1/(2n)))⇒(3/4), ((11)/(20)),((15)/(28)),...→(1/2)⇒inf=(1/2)  &  sup=(3/4)   (iii)  for the “n=2,4,8,10,14,..”f_n =−(1/2)(1+(1/(2n)))⇒−(5/8), −(9/(16)),−((17)/(32)),−((29)/(56))...→−(1/2)⇒inf=−(5/8)  &  sup=−(1/2)   (iv)  (i),(ii),(iii),(iv)⇒inf=−(7/6) &  sup=((13)/(12))

a)0,14,13,38,...12inf=0&sup=12b)forthen=odd13,37,511,...inf=12&sup=13(i)forthen=even25,49,613,...inf=25&sup=12(ii)(i),(ii)inf=12&sup=12c)forthen=odd0,0,0,...inf=sup=0forthen=even23,23,23,...inf=sup=23inf=0&sup=23d)1,0,1,0,1,0,1,...inf=1&sup=1e)forthen=evenen=1n12,14,16,...0inf=0&sup=12(i)forthen=4k+1en=1n10,45,89...1inf=1&sup=0(ii)forthen=4k+3en=1n+143,87,1211...1inf=1&sup=43(iii)(i),(ii),(iii)inf=1&sup=43f)forthen=6kfn=1+12n1312,2524,3736,...1inf=1&sup=1312(i)forthen=6k3fn=(1+12n)76,1918,3130,...1inf=76&sup=1(ii)forthen=1,5,7,11,13,..fn=12(1+12n)34,1120,1528,...12inf=12&sup=34(iii)forthen=2,4,8,10,14,..fn=12(1+12n)58,916,1732,2956...12inf=58&sup=12(iv)(i),(ii),(iii),(iv)inf=76&sup=1312

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