CSIR NET MATHEMATICS | Solution of Sequence & Series Example ( Using theorems and Counter examples)

In this video, we are going to discuss the example of sequence and series from CSINET June 2018. Example, if XN is a convergent sequence and YN bounded sequence in R, then we can conclude that first option XN plus YN is convergent, second option XN plus YN is bounded, third option XN plus YN has no convergent subsequence, fourth option XN plus YN has no bounded subsequence. We have theorem every convergent sequence is bounded and we have given that XN is convergent sequence so by this theorem XN is bounded sequence. One more theorem sum of two bounded sequences is bounded sequence since yn is bounded sequence therefore XN plus YN is a sum of two bounded sequence therefore it is also bounded and which is the second option therefore the second option Now we are going to check the other options. Let XN is equal to 1 and YN equal to minus 1 power n. XN equal to 1 is a constant sequence, therefore it is a convergent sequence. And YN equal to minus 1 power n is a bounded sequence which is bounded by minus 1 and plus 1. Now sum of XN plus YN is equal to 1 plus minus 1 power n. So it converts into this 0, 2, 0, 2, 0, 2 sequence which is divergent sequence. Therefore, the first option that XN plus YN is convergent, which is wrong. Now, this sequence XN plus YN has subsequence 2, 2, 2, 2, which is convergent as well as bounded subsequence. Therefore, the third option Xn plus Yn has no convergent subsequence is wrong. And Xn plus Yn has no bounded subsequence, which is also wrong. So, only second option is correct, Xn plus Yn is bounded if Xn is convergent and Yn is bounded sequence. Thanks for watching. For more video, like, subscribe, press the bell icon of NextGenMaths.