CSIR NET MATHEMATICS | Real Analysis | Limsup, Liminf, Convergence of sequence June 2019

Hello friends, in this video we are going to discuss the problem of real analysis from CSIR NET June 2019. Example number 30 from Booklet C, Marks 3. Consider a sequence a n, a n equal to minus 1 power n 1 by 2 minus 1 by n. Let b n equal to sigma k equal to 1 to n a k for all n belongs to n. Then which of the following is true? option limit n tends to infinity bn equal to 0 second option limit supremum n tends to infinity bn greater than 1 by 2 third option lim inf n tends to infinity bn less than minus 1 by 2 and the fourth option 0 less than equal to lim inf bn less than equal to lim sup bn less than equal to 1 by 2 so once we recall important result of sequence if a sequence of real numbers converges to real number x then any subsequence of xn also converges to x. In other words if a sequence xn of real numbers has two subsequence that converge to different limits then it diverges. That means sequence xn is diverges if we are able to find two subsequence which converge to different limits. So given sequence an is equal to minus 1 power n 1 by 2 minus 1 by n so first term of the sequence is n equal to 1 so minus 1 power 1 1 by 2 minus 1 by 1 so we have minus 1 by 2 plus 1 now a2 so for n equal to 2 minus 1 power 2 1 by 2 minus 1 by 2 now n equal to 3 so minus 1 power 3 1 by 2 minus 1 by 3 so we have minus 1 by 2 plus 1 by 3 a4 1 by 2 minus 1 by 4 a5 minus 1 by 2 plus 1 by 5 a6 1 by 2 minus 1 by 6 and the remaining terms now bn is equal to sigma k equal to 1 to n ak therefore b1 is equal to a1 and a1 has value 1 minus 1 by 2 b2 is sum of first two sequence of a and a1 plus a2 so if we take the summation of a1 plus a2 so minus 1 by 2 minus plus 1 by 2 cancel then we have remaining part 1 minus 1 by 2 b3 sum of first three terms of sequence a1 plus a2 plus a3 so minus 1 by 2 cancel with the plus 1 by 2 and the remaining part 1 minus 1 by 2 plus 1 by 3 and minus 1 by 2 similarly b4 so in b4 minus 1 by 2 cancel with the plus 1 by 2 1 minus 1 by 2 cancel with the plus 1 by 2 then we have 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 similarly b5 so we have minus 1 by 2 plus 1 by 2 cancel 1 more minus 1 by 2 plus 1 by 2 cancel and the remaining part 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 plus 1 by 5 and minus 1 by 2 similarly b6 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 plus 1 by 5 minus 1 by 6 so if you observe the terms than for even terms just like b2 we have 1 minus 1 by 2 for b4 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 for b6 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 plus 1 by 5 minus 1 by 6 and for odd terms we have 1 minus 1 by 2 for b3 1 minus 1 by 2 plus 1 by 3 and one more term which is minus 1 by 4 similarly b5 we have one more term minus 1 by 2 so if we generalize then bn is equal to 1 minus 1 by 2 up to 1 by n minus 1 by 2 if n is odd and if n is even then we have 1 minus 1 by 2 up to 1 upon n now if we recall this series of log 1 plus x equal to x minus x square by 2 plus x cube by 3 minus x power 4 by 4. If we take x is equal to 1 then log 1 plus 1 is equal to log 2 and 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 up to infinity terms. If n is odd then we have odd term sequence b1 b3 b5 b7 and if limit n tends to infinity b2n plus 1 approach to this bracket approach to log 2 and minus 1 by 2 is constant so it has limit log 2 minus 1 by 2. If n is even then the even term 2 4 8 converts to log 2. So we are able to find two subsequence one is odd term sequence second term is even term sequence have two different limits therefore by the theorem bn has two different sequence which converges to different limits and therefore it is diverges. so the first option limit n tends to infinity bn equal to 0 which is wrong now we recall b1 equal to 1 minus 1 by 2 so which has positive value b2 equal to a1 plus a2 which is also 1 minus 1 by 2 positive b3 1 minus 1 by 2 plus 1 by 3 minus 1 by 2 so you can see 1 minus 1 by 2 is plus 1 by 2 cancel the minus 1 by 2 so it has value 1 by 3 so it is also positive b4 1 minus 1 by 2 positive and 1 by 3 minus 1 by 4 is positive so sum of positive number is positive similarly b5 positive b6 positive so for all n bn is positive so lim inf bn which says is less than minus 1 by 2 so third option is wrong now as we know bn tends to infinity limit n tends to infinity b to n plus 1 has value log 2 minus 1 by 2 so the log 2 has value 0.6931 minus 0.5 which has value 0.1931 and log 2 has value 0.6931 so lim sup represent the maximum value or largest value of the sequence and in the fourth option we have given that lim sup bn less than equal to 1 by 2 but we are able to find the value which is bigger than 1 by 2 which is 0.6931 so the fourth option says that it is less than 1 by 2 which is wrong and the second option which says that lim sup is greater than 1 by 2 it is correct so thanks for watching for more videos like share subscribe press the bell icon of next yeah