Let's solve part **(b)** step by step: --- ### (b) Let {aₙ} and {bₙ} be sequences of real numbers. Suppose there exists n₀ ∈ ℕ such that aₙ ≤ bₙ for all n ≥ n₀. Prove that \[ \liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n \] and \[ \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n \] --- #### **Step 1: Prove for limsup** **Explanation:** Recall that: \[ \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup \{a_k : k \geq n\} \] \[ \limsup_{n \to \infty} b_n = \lim_{n \to \infty} \sup \{b_k : k \geq n\} \] Given: For all \( n \geq n_0 \), \( a_n \leq b_n \). Thus, for any fixed \( n \geq n_0 \), every term \( a_k \) with \( k \geq n \) is less than or equal to \( b_k \). Therefore: \[ \sup\{a_k : k \geq n\} \leq \sup\{b_k : k \geq n\} \] Taking the limit as \( n \to \infty \): \[ \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n \] --- #### **Step 2: Prove for liminf** **Explanation:** Recall that: \[ \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf \{a_k : k \geq n\} \] \[ \liminf_{n \to \infty} b_n = \lim_{n \to \infty} \inf \{b_k : k \geq n\} \] Similarly, for any fixed \( n \geq n_0 \), every term \( a_k \leq b_k \) for \( k \geq n \). So: \[ \inf\{a_k : k \geq n\} \leq \inf\{b_k : k \geq n\} \] Taking the limit as \( n \to \infty \): \[ \liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n \] --- ### **Final Answer:** If \( a_n \leq b_n \) for all \( n \geq n_0 \), then: \[ \boxed{ \liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n \quad \text{and} \quad \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n } \]

Let's approach part **(b)** carefully, breaking it down into two steps with clear reasoning. --- ### **Step 1: Show that \(\limsup a_n \leq \limsup b_n\)** **Explanation:** Given that there exists an \( n_0 \in \mathbb{N} \) such that for all \( n \geq n_0 \), \[ a_n \leq b_n \] Recall that: \[ \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k \] Similarly for \(b_n\): \[ \limsup_{n \to \infty} b_n = \lim_{n \to \infty} \sup_{k \geq n} b_k \] For any fixed \( n \geq n_0 \), since \( a_k \leq b_k \) for all \( k \geq n \), it follows that: \[ \sup_{k \geq n} a_k \leq \sup_{k \geq n} b_k \] As \( n \to \infty \), the limits of these supremums exist (possibly infinite), and thus: \[ \lim_{n \to \infty} \sup_{k \geq n} a_k \leq \lim_{n \to \infty} \sup_{k \geq n} b_k \] which implies: \[ \boxed{ \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n } \] --- ### **Step 2: Show that \(\liminf a_n \leq \liminf b_n\)** **Explanation:** Similarly, recall: \[ \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_k \] and \[ \liminf_{n \to \infty} b_n = \lim_{n \to \infty} \inf_{k \geq n} b_k \] For all \( n \geq n_0 \), since \(a_k \leq b_k\) for all \(k \geq n\), it follows that: \[ \inf_{k \geq n} a_k \leq \inf_{k \geq n} b_k \] Taking the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \inf_{k \geq n} a_k \leq \lim_{n \to \infty} \inf_{k \geq n} b_k \] which gives: \[ \boxed{ \liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n } \] --- ### **Final answer:** **Therefore,** \[ \boxed{ \text{If } a_n \leq b_n \text{ for all sufficiently large } n, \text{ then } \quad \liminf_{n \to \infty} a_n \leq \liminf_{n \to \infty} b_n \quad \text{and} \quad \limsup_{n \to \infty} a_n \leq \limsup_{n \to \infty} b_n } \]

Let's analyze part **(c)** step by step. --- ### (c) Sequences \(\{a_n\}\) and \(\{b_n\}\) are bounded below. Show that \[ \limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n \] --- ### **Step 1: Recall the definition of \(\limsup\) and properties of supremum** **Explanation:** By definition: \[ \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k \] and similarly for \(b_n\). We also know that for any sequences of real numbers, the supremum of the sum is less than or equal to the sum of the suprema: \[ \sup_{k \geq n} (a_k + b_k) \leq \sup_{k \geq n} a_k + \sup_{k \geq n} b_k \] **Note:** Since the sequences are bounded below, the supremums exist and are finite (possibly infinite, but for real sequences, we usually consider finite bounds). --- ### **Step 2: Use the inequality involving supremums to relate \(\limsup\) of sums** **Explanation:** For each \( n \), \[ \sup_{k \geq n} (a_k + b_k) \leq \sup_{k \geq n} a_k + \sup_{k \geq n} b_k \] Taking the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \sup_{k \geq n} (a_k + b_k) \leq \lim_{n \to \infty} \left( \sup_{k \geq n} a_k + \sup_{k \geq n} b_k \right) \] Since limits preserve addition when they exist (or are finite), and because the supremum sequences are bounded (by the assumption), we get: \[ \limsup_{n \to \infty} (a_n + b_n) \leq \left( \lim_{n \to \infty} \sup_{k \geq n} a_k \right) + \left( \lim_{n \to \infty} \sup_{k \geq n} b_k \right) = \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n \] --- ### **Final answer:** \[ \boxed{ \limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n } \] --- ### **Summary of reasoning:** - The key is the inequality \(\sup (a_k + b_k) \leq \sup a_k + \sup b_k\) for each \(n\). - Taking limits as \(n \to \infty\) preserves this inequality, leading to the desired result. --- Would you like to proceed with the remaining parts?

Let's proceed step by step with part **(d)**: --- ### (d) Given a sequence \(\{a_n\}\) of real numbers bounded above, define \(u_n = \sup\{a_k \mid k > n\}\). Show that: **(i)** \(\{u_n\}\) is a decreasing sequence. **(ii)** \(\limsup_{n \to \infty} a_n = \lim_{n \to \infty} u_n\). --- ### **Step 1: Show that \(\{u_n\}\) is decreasing** **Explanation:** By definition: \[ u_n = \sup_{k > n} a_k \] For each \( n \), the set \(\{a_k \mid k > n\}\) is contained within \(\{a_k \mid k > n-1\}\), since: \[ \{a_k \mid k > n\} \subseteq \{a_k \mid k > n-1\} \] Therefore, their supremums satisfy: \[ u_n = \sup_{k > n} a_k \leq \sup_{k > n-1} a_k = u_{n-1} \] Thus, the sequence \(\{u_n\}\) is **non-increasing** (decreasing or constant). Since the sequence is bounded above (by the initial sequence's bounds), it's decreasing and bounded below, so it converges. --- ### **Step 2: Show that \(\lim_{n \to \infty} u_n = \limsup_{n \to \infty} a_n\)** **Explanation:** Recall the definition: \[ \limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_k \] Note that: \[ u_n = \sup_{k > n} a_k \] and \[ \sup_{k \geq n} a_k \] are related, with the only difference being whether the supremum is over \(k > n\) or \(k \geq n\). Since: \[ \sup_{k > n} a_k \leq \sup_{k \geq n} a_k \] and as \( n \to \infty \), the difference between \(k > n\) and \(k \geq n\) becomes negligible, so: \[ \lim_{n \to \infty} u_n = \lim_{n \to \infty} \sup_{k > n} a_k = \lim_{n \to \infty} \sup_{k \geq n} a_k = \limsup_{n \to \infty} a_n \] --- ### **Final answers:** **(i)** \(\bullet\) \(\{u_n\}\) is decreasing because for all \( n \), \[ u_n \leq u_{n-1} \] **(ii)** \(\bullet\) \[ \boxed{ \lim_{n \to \infty} u_n = \limsup_{n \to \infty} a_n } \] --- **Summary:** The sequence \(u_n = \sup_{k > n} a_k\) captures the tail supremum and converges to the \(\limsup\) of the sequence \(a_n\). The decreasing nature follows from the nestedness of tail sets. Would you like to proceed with part **(e)**?

Let's analyze part **(e)** step by step. --- ### (e) Formulate (and prove) a result for \(\liminf a_n\) similar to part (d) --- ### **Step 1: Define the sequence \(l_n\)** **Explanation:** In analogy to \(u_n = \sup_{k > n} a_k\), for \(\liminf a_n\), we consider the sequence: \[ l_n = \inf_{k > n} a_k \] This sequence represents the infimum of the tail of \(\{a_k\}\) for \(k > n\). --- ### **Step 2: Show that \(\{l_n\}\) is increasing** **Explanation:** Since for each \( n \), \[ l_n = \inf_{k > n} a_k \] and the set \(\{a_k \mid k > n\}\) is contained in \(\{a_k \mid k > n-1\}\), we have: \[ \inf_{k > n} a_k \geq \inf_{k > n-1} a_k \] Hence, \[ l_n \geq l_{n-1} \] which shows that \(\{l_n\}\) is **non-decreasing**. --- ### **Step 3: Show that \(\lim_{n \to \infty} l_n = \liminf a_n\)** **Explanation:** Recall the definition of \(\liminf a_n\): \[ \liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_k \] Note that: \[ l_n = \inf_{k > n} a_k \] and \[ \inf_{k \geq n} a_k \] are close, with the only difference being whether the index is strictly greater than \( n \) or greater than or equal to \( n \). Since: \[ \inf_{k > n} a_k \geq \inf_{k \geq n} a_k \] and as \( n \to \infty \), the difference diminishes, we get: \[ \lim_{n \to \infty} l_n = \lim_{n \to \infty} \inf_{k > n} a_k = \lim_{n \to \infty} \inf_{k \geq n} a_k = \liminf_{n \to \infty} a_n \] --- ### **Final statement:** \[ \boxed{ \textbf{(e)} \quad \text{Define} \quad l_n = \inf_{k > n} a_k. } \] - The sequence \(\{l_n\}\) is **non-decreasing**. - \(\displaystyle \lim_{n \to \infty} l_n = \liminf_{n \to \infty} a_n\). --- ### **Summary:** - The sequence \(l_n = \inf_{k > n} a_k\) captures the tail infimum. - It is increasing, approaching the \(\liminf a_n\). Would you like to proceed with part **(f)**?

Let's analyze part **(f)** step-by-step. --- ### (f) Find \(\limsup a_n\) and \(\liminf a_n\) for the sequences: (i) \(a_n = (-1)^n (1 + 3/n)\) (ii) \(a_n = n^2 \sin^2(n)\) (iii) \(a_n = \lfloor n - 3 \rfloor\) --- ### **(i) \(a_n = (-1)^n (1 + 3/n)\)** --- ### **Step 1: Behavior of the sequence** - When \(n\) is even, \(a_n = (1 + 3/n)\) - When \(n\) is odd, \(a_n = - (1 + 3/n)\) As \(n \to \infty\): - \(a_n \to 1\) if \(n\) even (since \(3/n \to 0\)) - \(a_n \to -1\) if \(n\) odd --- ### **Step 2: Find \(\limsup a_n\) and \(\liminf a_n\)** - The subsequence with even \(n\): \(a_{2k} = 1 + 3/(2k) \to 1\) - The subsequence with odd \(n\): \(a_{2k+1} = - (1 + 3/(2k+1)) \to -1\) - **\(\limsup a_n\)** is the supremum of the subsequential limits: \[ \boxed{ \limsup a_n = 1 } \] - **\(\liminf a_n\)** is the infimum of the subsequential limits: \[ \boxed{ \liminf a_n = -1 } \] --- ### **Final answer:** \[ \boxed{ \limsup_{n \to \infty} a_n = 1, \quad \liminf_{n \to \infty} a_n = -1 } \] --- ### **(ii) \(a_n = n^2 \sin^2(n)\)** --- ### **Step 1: Behavior of the sequence** - Since \(\sin^2(n)\) oscillates between 0 and 1, \(a_n\) oscillates between: \[ 0 \quad \text{and} \quad n^2 \] - For values of \(n\) where \(\sin(n)\) is close to 0, \(a_n\) is close to 0 - For values of \(n\) where \(\sin(n)\) is close to \(\pm 1\), \(a_n\) is close to \(n^2\) --- ### **Step 2: Find \(\limsup a_n\) and \(\liminf a_n\)** - **\(\limsup a_n\)**: Since \(\sin^2(n)\) can be arbitrarily close to 1 for some subsequence, \[ \boxed{ \limsup_{n \to \infty} a_n = \infty } \] - **\(\liminf a_n\)**: Since \(\sin^2(n) \geq 0\), and can be arbitrarily close to 0, \[ \boxed{ \liminf_{n \to \infty} a_n = 0 } \] --- ### **Final answer:** \[ \boxed{ \limsup_{n \to \infty} a_n = \infty, \quad \liminf_{n \to \infty} a_n = 0 } \] --- ### **(iii) \(a_n = \lfloor n - 3 \rfloor\)** --- ### **Step 1: Behavior of the sequence** - For large \(n\), \[ a_n = \lfloor n - 3 \rfloor \] - As \(n \to \infty\), \(a_n \to n - 3\), which diverges to infinity. --- ### **Step 2: Find \(\limsup a_n\) and \(\liminf a_n\)** - Since the sequence increases without bound, both \(\limsup\) and \(\liminf\) are infinite: \[ \boxed{ \limsup_{n \to \infty} a_n = \infty, \quad \liminf_{n \to \infty} a_n = \infty } \] --- ### **Final summarized results:** | Sequence | \(\limsup a_n\) | \(\liminf a_n\) | |------------|------------------|-----------------| | (i) \((-1)^n (1 + 3/n)\) | 1 | -1 | | (ii) \(n^2 \sin^2(n)\) | \(\infty\) | 0 | | (iii) \(\lfloor n -3 \rfloor\) | \(\infty\) | \(\infty\) | --- Would you like a recap or further clarification on any part?

Let's carefully approach part **(a)** step by step. --- ### **Part (a): Characterization of \(\liminf_{n \to \infty} a_n\)** --- ### **Step 1: Recall the characterization of \(\limsup_{n \to \infty} a_n\)** **Explanation:** From Theorem 03.6 (or standard results), the \(\limsup\) of a sequence \(\{a_n\}\) can be characterized as: \[ \boxed{ \limsup_{n \to \infty} a_n = \inf_{N \in \mathbb{N}} \sup_{n \geq N} a_n } \] This means the limit superior is the greatest lower bound (infimum) over all \(N\) of the supremums of the tail sequences. --- ### **Step 2: Seek a dual characterization for \(\liminf_{n \to \infty} a_n\)** **Explanation:** Since \(\liminf\) and \(\limsup\) are dual notions, their characterizations mirror each other, with \(\sup\) replaced by \(\inf\), and vice versa. By analogy, the characterization for \(\liminf_{n \to \infty} a_n\) should be: \[ \boxed{ \liminf_{n \to \infty} a_n = \sup_{N \in \mathbb{N}} \inf_{n \geq N} a_n } \] **Justification:** - For each fixed \(N\), \(\inf_{n \geq N} a_n\) gives the smallest value in the tail starting from \(N\). - Taking the supremum over all such \(N\) finds the largest value among these lower bounds, capturing the "best" lower limit point as \(N \to \infty\). --- ### **Step 3: **Prove the characterization** **Proof:** - **(i)** For any \(N \in \mathbb{N}\), \[ \inf_{n \geq N} a_n \leq a_n \quad \text{for all } n \geq N \] - **(ii)** As \(N \to \infty\), the sequence \(\{\inf_{n \geq N} a_n\}\) is **non-decreasing** (since as \(N\) increases, the tail gets smaller or stays the same). - **(iii)** The limit of these infima as \(N \to \infty\) is the \(\liminf a_n\): \[ \lim_{N \to \infty} \inf_{n \geq N} a_n = \liminf_{n \to \infty} a_n \] - **(iv)** The supremum over all \(N\) of the \(\inf_{n \geq N} a_n\) is the least upper bound of these tail infima, which equals the limit: \[ \boxed{ \liminf_{n \to \infty} a_n = \sup_{N \in \mathbb{N}} \inf_{n \geq N} a_n } \] since the sequence \(\{\inf_{n \geq N} a_n\}\) increases to \(\liminf a_n\). --- ### **Final answer:** \[ \boxed{ \textbf{A characterization of } \liminf_{n \to \infty} a_n \text{ is:} \quad \boxed{ \liminf_{n \to \infty} a_n = \sup_{N \in \mathbb{N}} \inf_{n \geq N} a_n } } \] --- ### **Summary:** - The \(\liminf\) of \(\{a_n\}\) can be viewed as the greatest lower bound, over all tails, of the sequence. - This dualizes the \(\limsup\) characterization, which involves infimums over tails. --- Would you like a further explanation or to connect this with the statement in Theorem 03.6?