ISI-2017-MMA-16

Question

Let $(x_n)$ be a sequence of a real number such that the subsequence $(x_{2n})$ and $(x_{3n})$ converge to limit $K$ and $L$ respectively. Then

• A. $(x_n)$ always converge
• B. If $K=L$ then $(x_n)$ converge
• C. $(x_n)$ may not converge but $K=L$
• D. it is possible to have $K \neq L$

Comments

option C

If k=l the sequence may not converge.

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When limit increasing if the number going towards 0, then it will converge

no options matching

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Srestha I am not getting what are u saying?

Answer 1

Consider the theorem 

If a sequence X_n is convergent to $ l $ then all its subsequences X_m is convergent to $ l $ .

Now consider the sequence 

                                    Xn = -1   when n is a prime number.

                           and   Xn =  1   otherwise.

Then we get that the subsequence X_2n is  1,1,1,........ thus the sequence X_2n is convergent to 1. 

Hence k=1.

Again we also get that X_3n is 1,1,1,........... thus the sequence X_3n is convergent to 1.

Hence $ l $ = 1.

So, k = $ l $.

Now, consider the subsequence X_m where m is a prime. Then X_m is -1,-1,-1,......... thus the sequence X_m is convergent to -1.

So, we get that all the subsequences of X_n is not convergent to a particular number. So, X_n is divergent. But here we have k = $ l $.

So, option C is correct X_n may not converge but k = $ l $.