CSIR NET MATHEMATICS | Real Analysis | Convergence of Series Dec 2017 (Marks 4.75)

Hello friends, in this video we are going to discuss the problem of re-analysis from CSI and NIT December 2017. Example from booklet A, marks 4.75. Question, which of the following are convergent? First option, sigma n equal to 1 to infinity n square 2 power minus n. Second option, sigma n equal to 1 to infinity n power minus 2 2 power n. Third option, sigma n equal to 2 to infinity 1 upon n log n. And the fourth option sigma n equal to 1 to infinity 1 upon So, first we recall some test of convergence of positive term series. Let sigma n equal to 1 to infinity be positive term series. First test, ratio test. Limit n tends to infinity, ratio a n plus 1 upon a n equal to l. If l less than 1, then series is convergent. l greater than 1, series is divergent. l equal to 1, then test is failed to check the convergence or divergence of series. Root test. n tends to infinity nth root of an equal to l if l is than 1 sigma an is convergent series greater than 1 then sigma an is divergent series equal to 1 then test is failed to check the convergence. Integral test in the integral test we assume an is equal to fn where f is a function of n if integration 1 to infinity fx dx improper integral is convergent Then this series is convergent otherwise series is divergent. Taste of divergent if limit n tends to infinity a n not equal to 0. If nth term has limit non-zero then series is divergent. So we will use this taste to check the convergence of series given in the options. First option we will use root taste where a n is equal to n square into 2 power minus n. nth root of a n n square into 2 power minus n power 1 by n. Then we have n square power 1 by n and minus n cancel with the 1 by n then 2 power minus 1. We can rewrite n square power 1 by n in the form of n power 1 by n power 2. Now, we have limit result limit n tends to infinity n power 1 by n equal to 1. Therefore, this limit has value 1 by 2 which is less than 1. Therefore, by the root test series is convergent. So, first option will give the convergent series. Now the second option where a n is equal to n power minus 2, 2 power n. So nth root of a n, n power minus 2, 2 power n power 1 by n. Then we have n power 1 by n power minus 2 into 2 where n cancel with 1 by n. And as we know limit n tends to infinity, n power 1 by n has value 1. Therefore this series has value 2 which is greater than 2. And hence by the root test sigma n equal to 1 to infinity n power minus 2, 2 power n is divergent. So, therefore the second option will give the divergence series. Third option where we will use integral test where a n is equal to 1 upon n log n. In the integral test a n equal to f of n equal to 1 upon n log n. Now, integration 2 to infinity f x dx because we will start from 2 to infinity. So, integration 2 to infinity 1 upon x log x. We can rewrite 1 upon x divided by log x. So, in the numerator we have derivative of function. function in the denominator we have function So f dash upon f using integration result we have the integration log of log x evaluate between 2 to infinity if you put x is equal to infinity then integration has value infinity log 2 will get the finite value so infinity minus finite value will always give the infinity therefore this integration has value infinity and hence this integration is divergent so the series sigma n equal to 2 to infinity 1 upon n log n is also divergent now The fourth option where we will use Divergent Taste, a n is equal to 1 upon n log 1 plus 1 by n. Just recall the series of log 1 plus x, x minus x square by 2 plus x cube by 3 minus x power 4 by 4. Now a n is equal to 1 upon n log 1 plus 1 upon n. Now here x replaced by 1 by n then we have series 1 by n minus x replaced by 1 by n then the second term is 1 upon 2 n square plus 1 upon 3 n cube minus 1 upon 4 n power 4. Multiply n in the bracket then n cancel with the n then we have first term 1 and the remaining terms minus 1 upon 2 n plus 1 upon 3 n square minus 1 upon 4 n cube. Now apply limit on the both side limit n tends to infinity a n is equal to limit n tends to infinity As n tends to infinity, 1 upon n tends to 0, so all the terms are 0 except the first terms and we have 1 upon 1 equal to 1, which is non-zero. Therefore, limit n tends to infinity, a n is non-zero, hence the series is divergent. Therefore, fourth option will also give the divergent series. So, there is only one convergent series, which is in the first option. Thanks for watching. For more videos, like, share, subscribe, press the bell icon of NextGenMaths.