The sequence given is $a_n = \frac{(-1)^n}{n}$, where $n \geq 1$. We are tasked with determining the limit of this sequence using the Squeeze Theorem.

Step 1: Understanding the Sequence

The sequence alternates signs because of the $(-1)^n$ term, which makes the terms alternate between positive and negative values. The magnitude of each term is $\frac{1}{n}$, which decreases as $n$ increases.

Step 2: Using the Squeeze Theorem

To apply the Squeeze Theorem, we need to "squeeze" the given sequence between two simpler sequences whose limits we know.

1. Bound the Sequence:

   • Notice that $-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}$. This is because $(-1)^n$ only affects the sign, not the magnitude, and the magnitude is $\frac{1}{n}$.

   Therefore, we have the inequality: $-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}$

2. Take Limits of the Boundaries:

   • As $n \to \infty$, both $-\frac{1}{n}$ and $\frac{1}{n}$ tend to 0. $\lim_{n \to \infty} -\frac{1}{n} = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{n} = 0$

3. Apply the Squeeze Theorem:

   • Since $-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}$ and both the lower and upper bounds tend to 0, by the Squeeze Theorem: $\lim_{n \to \infty} \frac{(-1)^n}{n} = 0$

Conclusion:

The limit of the sequence $a_n = \frac{(-1)^n}{n}$ as $n \to \infty$ is 0.

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. How is the Squeeze Theorem generally applied in other sequences?
2. Can you apply the Squeeze Theorem to sequences involving trigonometric functions?
3. What happens if the numerator grows faster than the denominator?
4. How can you use the Squeeze Theorem in multivariable calculus?
5. What is the difference between the Squeeze Theorem and L'Hopital's Rule in finding limits?

Tip: The Squeeze Theorem is especially useful for alternating sequences or cases where you can easily find upper and lower bounds that converge to the same limit.