CSIR NET MATHEMATICS | How to solve the problem of Sequence & Series? (Method of Counter Examples)

In this video we are going to discuss example of sequence and series from the CSINET June 2018. Example, given a n, b n two monotonic sequences of real numbers and that sigma a n into b n convergent which of the following is true. First option, sigma a n is convergent and sigma b n is convergent. Second option, at least one of sigma a n b n is convergent. Third option, sigma an is bounded and bn is bounded. At least one of an, bn is bounded. To answer this question, we will use method of counter example. For the option 1, an equal to 1 by n and bn is equal to 1 upon n square. Sequence 1 upon n is monotonic decreasing sequence and the series sigma 1 upon n is divergent by the p-test Sigma 1 upon n power p is convergent if p greater than 1 and divergent if p less than equal to 1. bn is equal to 1 upon n square which is monotonic decreasing sequence and the series sigma 1 upon n square is convergent by the same p test because p equal to 2 which is greater than 1. Now sigma an into bn is equal to 1 upon n into 1 upon n square. So we have 1 upon n cube which is convergent This is very important because p equal to 3 in the p-test. Now, we can observe that the sigma an is divergent, sigma bn is convergent and sigma an into bn is convergent. That means sigma an into bn is convergent even though an is divergent. So, therefore, the first option sigma an is convergent and bn is convergent which is wrong. Now, the second option. For the second option, we assume an equal to 1 upon n, bn equal to 1 upon n. an equal to bn equal to 1 by n are monotonic decreasing sequence. And the series sigma a n equal to b n equal to 1 by n are divergent series by the p test as p equal to 1. Now the multiplication sigma a n into b n is equal to 1 upon n into 1 upon n equal to 1 upon n square p equal to 2. So therefore it is a convergent sequence. So, now here we can observe that AN and BN are divergent. Sigma an into bn is convergent and option at least one of sigma an bn is convergent which is also not correct. Now option three for option three we assume an equal to n and bn equal to one upon n cube. So an is monotonic increasing sequence but not bounded which is bounded below below by one and the second sequence bn equal to one upon n cube is monotonic decreasing sequence and bounded which is bounded by 0 and 1 then the product sigma a n b n equal to 1 n into 1 upon n cube is equal to 1 upon n square which is convergent. So here we can observe that sigma a n is unbounded and b n is bounded and sigma a n into b n 1 upon n square is convergent. So third option that sigma a n is bounded and sigma b n is bounded which is also wrong. So, therefore the fourth option that means at least one of a n and b n is bounded which is correct and we have one example that here b n equal to one upon n cube is bounded. Thanks for watching. For more videos like, subscribe, press the bell icon of next gen.