This is a problem based on the Squeeze Theorem (or Sandwich Theorem) for sequences.

Problem:

We have three sequences $\{a_n\}_{n=1}^\infty$, $\{b_n\}_{n=1}^\infty$, and $\{c_n\}_{n=1}^\infty$ with the following conditions:

1. $\{a_n\}$ converges to $A$.
2. $\{b_n\}$ converges to $A$.
3. For all $n$, $a_n \leq c_n \leq b_n$.

The task is to prove that $\{c_n\}$ also converges to $A$.

Solution Outline:

The Squeeze Theorem states that if $\{a_n\} \rightarrow A$ and $\{b_n\} \rightarrow A$, and if $a_n \leq c_n \leq b_n$ for all $n$, then $\{c_n\} \rightarrow A$ as well.

1. Since $\{a_n\} \rightarrow A$, for any $\epsilon > 0$, there exists an $N_1$ such that for all $n \geq N_1$, $|a_n - A| < \epsilon$.

2. Similarly, since $\{b_n\} \rightarrow A$, for any $\epsilon > 0$, there exists an $N_2$ such that for all $n \geq N_2$, $|b_n - A| < \epsilon$.

3. Let $N = \max(N_1, N_2)$. Then for all $n \geq N$, both $|a_n - A| < \epsilon$ and $|b_n - A| < \epsilon$.

4. Since $a_n \leq c_n \leq b_n$ for all $n$, it follows that: $-\epsilon < a_n - A \leq c_n - A \leq b_n - A < \epsilon.$ This implies that $|c_n - A| < \epsilon$ for all $n \geq N$.

5. Therefore, by the definition of convergence, $\{c_n\}$ converges to $A$.

This completes the proof that $\{c_n\} \rightarrow A$.

Would you like further details on any step, or have any questions?

Related Questions

1. How is the Squeeze Theorem applied in calculus beyond sequences?
2. What are some examples where the Squeeze Theorem is useful for functions?
3. Can the Squeeze Theorem be applied if only one of the bounding sequences converges?
4. What happens if $a_n$ and $b_n$ converge to different limits?
5. How would this proof change if we were dealing with functions instead of sequences?

Tip

When using the Squeeze Theorem, always ensure that the bounding sequences or functions converge to the same limit. This is essential for the theorem to apply correctly.