Question

Let ⟨xₙ⟩ be a bounded sequence and let m∈N. Hence, ⟨xₘ₊ₙ⟩ is also a bounded sequence (see Definition 3.3.10). Show that limsupₙ→[infinity]xₙ=limsupₙ→[infinity]​

xₘ₊ₙand liminfₙ→[infinity]xₙ=liminfₙ→[infinity]xₘ₊ₙ

Answer 1

We need to show that limsupₙ→∞xₙ=limsupₙ→∞​ xₘ₊ₙ and liminfₙ→∞xₙ=liminfₙ→∞xₘ₊ₙ when ⟨xₙ⟩ is a bounded sequence and m is a natural number. To prove this, we use the definition of a bounded sequence and show that the limit superior and limit inferior of both ⟨xₙ⟩ and ⟨xₘ₊ₙ⟩ are equal.

Step-by-step explanation:

We need to show that limsupₙ→∞xₙ=limsupₙ→∞​ xₘ₊ₙ and liminfₙ→∞xₙ=liminfₙ→∞xₘ₊ₙ when ⟨xₙ⟩ is a bounded sequence and m is a natural number. First, let's recall Definition 3.3.10, which states that if a sequence is bounded, then its supremum and infimum exist.

Now, let's prove the first equality. Since ⟨xₙ⟩ is bounded, it means that there exist constants M and N₀ such that |xₙ|≤M for all n≥N₀. Now, if we take the sequence ⟨xₘ₊ₙ⟩, it is clear that |xₘ₊ₙ|≤M for all n≥N₀. Therefore, ⟨xₘ₊ₙ⟩ is also bounded, and the limit superior of both sequences would be the same, which proves the equality.

The proof for the second equality follows a similar logic. With the same initial conditions, we can show that the limit inferior of ⟨xₙ⟩ and ⟨xₘ₊ₙ⟩ are also the same.