Final Answer:
The sequence (-1)ⁿₙ∈ℕ is divergent. This can be proven by considering two subsequences: one formed by the even indices, which converges to 1, and another formed by the odd indices, which converges to -1, indicating that the original sequence does not have a single limit and is therefore divergent.
Step-by-step explanation:
To establish the divergence of the sequence (-1)ⁿₙ∈ℕ, we can examine the behavior of its subsequences. Let's consider two subsequences: one formed by taking only the terms with even indices (n = 2k, where k is a non-negative integer) and another formed by taking only the terms with odd indices (n = 2k + 1). The even subsequence is given by (-1)²ⁿ = 1 and converges to 1, while the odd subsequence is given by (-1)²ⁿ⁺¹ = -1 and converges to -1.
As these subsequences have different limits, the original sequence (-1)ⁿₙ∈ℕ does not converge to a single value. By the definition of a divergent sequence, if there exist two subsequences with distinct limits, the original sequence is considered divergent. In this case, the even and odd subsequences demonstrate that the original sequence does not have a convergent limit, proving its divergence.
This divergence is a characteristic feature of oscillating sequences, where the terms alternate between different values without approaching a specific limit. The use of convergency of subsequences provides a concise and rigorous way to demonstrate the divergence of the given sequence.