91Ó°ÊÓ

To prove or provide a counterexample for this statement, we first need to recall two definitions. A sequence \(s_n\) is Cauchy if for every \(\epsilon > 0\), there exists an integer \(M\) such that for all the terms \(s_m\) and \(s_n\) with \(m, n > M\), the absolute difference between these terms is less than \(\epsilon\), in mathematical terms: \[|s_m - s_n| < \epsilon\] A sequence \(s_n\) is bounded if there exist two real numbers \(L\) and \(U\) such that for all \(n\), \(L \leq s_n \leq U\). We can use the Bolzano-Weierstrass theorem, which states that every bounded sequence has a convergent subsequence. And, since a convergent subsequence is a Cauchy sequence, thus we can conclude that every bounded sequence has a Cauchy subsequence. The statement is true.

To prove or provide a counterexample for this statement, we first need to recall the definition of a monotone sequence and a bounded sequence. A sequence \(s_n\) is called monotone if either of the following conditions is true: - The sequence is strictly increasing: \(s_{n+1} > s_n\) for all \(n\) - The sequence is strictly decreasing: \(s_{n+1} < s_n\) for all \(n\) Now, if a monotone sequence is bounded, either from the below (for increasing sequences) or from above (for decreasing sequences), it has a convergent subsequence (by Bolzano-Weierstrass theorem). And since the original monotone sequence is either strictly increasing or strictly decreasing, the convergent subsequence will share the monotonicity with the original sequence. Therefore, the statement is false, as it is not possible to find a bounded subsequence.

To prove or provide a counterexample for this statement, we first need to recall the definition of a convergent sequence and an oscillating sequence. A sequence \(s_n\) is called convergent if there exists a limit point \(L\) such that as \(n\) approaches infinity, the term \(s_n\) approaches the limit point: \[\lim_{n\to\infty} s_n = L\] A sequence \(s_n\) is called oscillating if it alternates between increasing and decreasing, with no converging behavior. Now, let's consider a convergent sequence \(s_n\), with the limit point \(L\). We can represent the sequence as the sum of two oscillating sequences \(a_n\) and \(b_n\) by defining them as follows: - \(a_n = \frac{s_n + s_{n+1}}{2}\) - \(b_n = \frac{s_n - s_{n+1}}{2}\) Both of the sequences oscillate and their sum is \(s_n = a_n + b_n\). Hence, every convergent sequence can be represented as the sum of two oscillating sequences, and the statement is true.