Here are two useful definitions: A sequence (a_n) is eventually in a set A R if

Question

Here are two useful definitions: A sequence (a_n) is eventually in a set A R if there exists an N N such that a_n A for all n N. A sequence (a_n) is frequently in a set A R if for every N N, there exists an n N such that a_n A. Is the sequence ( - 1)^n eventually or frequently in the set {1}. Which definition is stronger? Does frequently imply eventually? Does eventually imply frequently? Given an alternative rephrasing of Definition 2.2.3B using either frequently or eventually. Suppose an infinite number of terms of a sequence (x_n) are equal to 2. Is (x_n) necessarily eventually in the interval (1.9,2.1)? Is it frequently in(1.9,2.1)?

Explanation / Answer

Solution:

(a) The terms of (a_n) are (-1, 1, -1, 1....). a_n = -1 if n is odd and a_n = -1 if n is even. So, for infinitely many n's, a_n = 1 and, also for infinitely many a_n = - 1.

    There's no N such that a _n is in {1} for every n N. But for every N there is n N with a_n in {1}.

    (a_n) is frequently in {1} and also in {-1}. And eventually in {-1, 1}.

b) It's clear that eventually implies frequently.

   If thee is N such that a_n is A for n N, then for every K there is n K with a_n in A. Just choose n > max{N, K}.
   The converse is not true. The sequence in (a) shows this.

c) what is definition 2.2.3B? Anyway, the definitions you gave are equivalent to:

   (a_n) is eventually in A if a_n is in A for all but a finite number of indexes n.

   (a_n) is frequently in A if a_n is in A for infinitely many indexes n.


d) No for eventually and yes for frequently. x_n = 2 (-1)^n is frequently in this interval but not eventually in it.

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