asked 192k views
5 votes
The sequence (an) is not bounded above if and only if there exists a subsequence (ank) that _____.

A. Converges.
B. Diverges.
C. Is constant.
D. Goes to infinity.

asked
User Sveisvei
by
6.9k points

1 Answer

4 votes

Final answer:

The sequence (an) is not bounded above if and only if there exists a subsequence (ank) that goes to infinity. This means that the terms in the subsequence become larger and larger as the index, k, increases. Therefore, the correct answer is D. Goes to infinity.

Step-by-step explanation:

The sequence (an) is not bounded above if and only if there exists a subsequence (ank) that goes to infinity. This means that the terms in the subsequence become larger and larger as the index, k, increases.

To illustrate this, let's consider an example. Suppose our original sequence (an) is given by an = n, where n is a positive integer. This sequence is not bounded above because as n increases, an also increases without bound. Now, we can create a subsequence (ank) where k is even, such that ank = 2k. This subsequence goes to infinity because as k increases, ank becomes larger and larger.

Therefore, the correct answer is D. Goes to infinity.

answered
User Loganasherjones
by
7.8k points
Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.