Question

How would you show a sequence unbounded from above has a subsequence with limit infinity?

A. By using the Bolzano-Weierstrass theorem.

B. By demonstrating that the sequence is monotonic.

C. By showing that for any M>0, there exists an N such that s_N>M.

D. By proving that the sequence is convergent.

Answer 1

To show that a sequence unbounded from above has a subsequence with limit infinity, we need to demonstrate that for any positive value M, there is a term in the sequence that is larger than M.

Step-by-step explanation:

The correct answer is C. By showing that for any M>0, there exists an N such that sN>M.

To show that a sequence unbounded from above has a subsequence with limit infinity, we need to demonstrate that for any positive value M, there is a term in the sequence that is larger than M. This implies that the elements of the sequence can get arbitrarily large, and thus, the sequence has a subsequence that tends to infinity.

For example, suppose we have a sequence {sn} that is unbounded from above. This means that for any positive number M, there exists an index N such that sN > M. Thus, we can choose M to be any positive number, and we can always find an index N such that sN is greater than M. Therefore, the sequence has a subsequence that tends to infinity.