Problem 2. Let {an}n≥1 be a sequence. Prove that if {akn }n≥1 is a subsequence of {an}n≥1that converges to L ∈ R, thenlim infn→∞ an ≤ L ≤ lim supn→∞an.(Hint: As {an}n≥1 is not necessarily bounded, lim infn→∞ an and lim supn→∞an might be equal to±∞ here. However, in these cases, the proof is much easier.)
Question
Problem 2. Let {an}n≥1 be a sequence. Prove that if {akn }n≥1 is a subsequence of {an}n≥1that converges to L ∈ R, thenlim infn→∞ an ≤ L ≤ lim supn→∞an.(Hint: As {an}n≥1 is not necessarily bounded, lim infn→∞ an and lim supn→∞an might be equal to±∞ here. However, in these cases, the proof is much easier.)
Solution
To prove this, we need to understand the definitions of limit superior (lim sup) and limit inferior (lim inf).
The limit superior of a sequence {an} as n approaches infinity is the smallest number L such that, for any positive number ε, there is a number N such that for all n > N, an < L + ε. In other words, beyond some point in the sequence, all terms are less than L + ε.
The limit inferior of a sequence {an} as n approaches infinity is the largest number L such that, for any positive number ε, there is a number N such that for all n > N, an > L - ε. In other words, beyond some point in the sequence, all terms are greater than L - ε.
Given that {akn} is a subsequence of {an} that converges to L, by definition of convergence, for any positive number ε, there is a number N such that for all n > N, akn is within ε of L.
Now, let's prove the two inequalities:
- lim inf an ≤ L:
Since {akn} is a subsequence of {an}, for any n, there exists k such that akn ≥ an. Therefore, for any ε > 0, there exists N such that for all n > N, akn > L - ε. Hence, an > L - ε for all n > N, which means L is an upper bound for the set {an: n > N}. Therefore, L ≥ lim inf an.
- L ≤ lim sup an:
Similarly, since {akn} is a subsequence of {an}, for any n, there exists k such that akn ≤ an. Therefore, for any ε > 0, there exists N such that for all n > N, akn < L + ε. Hence, an < L + ε for all n > N, which means L is a lower bound for the set {an: n > N}. Therefore, L ≤ lim sup an.
Therefore, we have proven that if {akn} is a subsequence of {an} that converges to L, then lim inf an ≤ L ≤ lim sup an.
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