Other Important Criteria

Idea: Use the subsequence criterion to find two subsequences converging to different limits.

Detailed steps:

1. Take the subsequence $\{x_{2n}\}$: $x_{2n} = (-1)^{2n} = 1$, whose limit is 1

2. Take the subsequence $\{x_{2n-1}\}$: $x_{2n-1} = (-1)^{2n-1} = -1$, whose limit is -1

3. Since there exist two subsequences converging to different limits, the original sequence diverges

Answer: The sequence diverges.

Idea: Use the boundedness criterion to prove that the sequence is unbounded.

Detailed steps:

1. For any positive number $M$, take $N = \lceil M \rceil + 1$

2. When $n > N$, $x_n = n > N > M$

3. Therefore the sequence is unbounded

4. By the contrapositive of the boundedness criterion, the sequence diverges

Answer: The sequence diverges.

Idea: Use the definition to prove that the sequence converges.

Detailed steps:

1. For any $\varepsilon > 0$, take $N = \left\lceil \frac{1}{\varepsilon} \right\rceil$

2. When $n > N$, $|x_n - 0| = \frac{1}{n} < \frac{1}{N} \leq \varepsilon$

3. Therefore $\lim x_n = 0$

Answer: The sequence converges to 0.

Idea: First find the limit, then use the sign-preserving property.

Detailed steps:

1. $\lim_{n \to \infty} \frac{n^2 + 1}{n^2 + n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n}} = 1 > 0$

2. By the sign-preserving property, there exists $N$ such that when $n > N$, $x_n > 0$

3. Therefore the sequence is positive from some term onward

Answer: The limit is positive, and the sequence is positive from some term onward.

Idea: Use the subsequence criterion to find subsequences converging to different limits.

Detailed steps:

1. Take the subsequence $\{x_{4n}\}$: $x_{4n} = \sin(2n\pi) = 0$, whose limit is 0

2. Take the subsequence $\{x_{4n+1}\}$: $x_{4n+1} = \sin(2n\pi + \frac{\pi}{2}) = 1$, whose limit is 1

3. Since there exist two subsequences converging to different limits, the original sequence diverges

Answer: The sequence diverges.