Which one of the following is TRUE ?a)Every sequence that has a conver...
Explanation:
Convergent Subsequence: A subsequence {a_nk} of a sequence {a_n} is said to be convergent if it converges to a limit L, i.e., if lim_{k-->infinity} a_nk = L.
Cauchy Sequence: A sequence {a_n} is said to be Cauchy if for every positive number epsilon, there exists a positive integer N such that, for all m,n ≥ N, |a_m - a_n| < />
Bounded Sequence: A sequence {a_n} is said to be bounded if there exists a positive number M such that |a_n| ≤ M for all n.
Now, let's consider each option:
a) Every sequence that has a convergent subsequence is a Cauchy sequence.
This statement is false. A counterexample is the sequence {a_n} = {1, 1/2, 2, 1/3, 3, 1/4, 4, ...}. The subsequence {1/n} is convergent (to 0), but the sequence is not Cauchy since |a_{2n} - a_n| = n > epsilon for any positive epsilon.
b) Every sequence that has a convergent subsequence is a bounded sequence.
This statement is also false. A counterexample is the sequence {a_n} = {n, n+1, n+2, ...}. The subsequence {n} is convergent (to infinity), but the sequence is not bounded.
c) The sequence {sin n} has a convergent subsequence.
This statement is true. The sequence {sin n} oscillates between -1 and 1. However, it has a subsequence {sin (2nπ)} that converges to 0 as n goes to infinity. Therefore, the statement is true.
d) The sequence {ncos1/n} has a convergent subsequence.
This statement is also true. Note that, as n goes to infinity, 1/n goes to 0, so cos(1/n) goes to 1. Therefore, the sequence {ncos1/n} goes to infinity, but it has a subsequence {n*cos(1/n^2)} that converges to 1 as n goes to infinity. Therefore, the statement is true.
Therefore, the correct answer is option c).