Question

For each n ∈ N, let Xn be the sequence Xn = ⟨xn,k : k ∈N⟩, and suppose that lim Xn = an and that lim an = a. Show that there exist positive integers k(n) (or kn if you prefer) such that lim xn,k(n) = a. note: k,n is the subscript of x (this was confusing for me at first)

Answer 1

To show that there exist positive integers k(n) such that lim xn,k(n) = a, we can use the definition of a limit. By choosing n = max(N, K), where N and K are positive integers satisfying the limit conditions, we can ensure that ||Xn - xn,k(n)|| < ϵ and |xn,k(n) - an| < δ are both satisfied. Therefore, we have shown the existence of k(n) such that lim xn,k(n) = a.

Step-by-step explanation:

In order to show that there exist positive integers k(n) such that lim xn,k(n) = a, we can use the definition of a limit. Since lim Xn = an, we know that for any positive number ϵ, there exists a positive integer N such that for all n > N, ||Xn - an|| < ϵ. Similarly, since lim an = a, for any positive number δ, there exists a positive integer K such that for all k > K, |ak - a| < δ. By choosing n = max(N, K), we can ensure that both ||Xn - an|| < ϵ and |an - a| < δ are satisfied. Now, let's define k(n) = max(N, K). Since n > N, we have k(n) > N, meaning ||Xn - xn,k(n)|| < ϵ. And since also k(n) > K, we have k(n) > K, and thus |xn,k(n) - an| < δ. Therefore, by choosing k(n) = max(N, K), we have shown that there exist positive integers k(n) such that lim xn,k(n) = a.