Let's go through each of the questions in the image one by one:

Problem 8:

If $(s_n)$ is a subsequence of $(t_n)$ and $(t_n)$ is a subsequence of $(s_n)$, can we conclude that $(s_n) = (t_n)$? Prove or give a counterexample.

Solution:

We need to decide if two sequences that are subsequences of each other must be identical. A counterexample would disprove this claim.

Counterexample: Consider the sequence $(s_n) = (-1)^n$, which alternates between $1$ and $-1$. Now, define:

• $(t_n)$ as the subsequence of $(s_n)$ where $t_n = s_{2n} = 1$ for all $n$, i.e., it takes only the positive terms from $(s_n)$.
• $(s_n)$ as the subsequence of $(t_n)$, where $s_n = t_{2n} = 1$, i.e., it also only includes positive terms.

Even though both are subsequences of each other, they are not identical because one sequence consists of both $1$ and $-1$, while the other only consists of $1$.

Thus, no, we cannot conclude that $(s_n) = (t_n)$ if they are subsequences of each other.

Problem 9:

Let $(s_n)$ be a bounded sequence and suppose that $\liminf s_n = \limsup s_n = s$. Prove that $(s_n)$ is convergent and that $\lim s_n = s$.

Solution:

We are given that the sequence $(s_n)$ is bounded, and $\liminf s_n = \limsup s_n = s$. We want to prove that the sequence converges to $s$.

1. Boundedness: Since $(s_n)$ is bounded, it means there exists some real numbers $m$ and $M$ such that $m \leq s_n \leq M$ for all $n$.

2. Equality of $\liminf$ and $\limsup$: The limit inferior ($\liminf s_n$) is the largest value such that any subsequence converging below it must converge to a limit smaller than or equal to this value. Similarly, the limit superior ($\limsup s_n$) is the smallest value such that any subsequence converging above it must converge to a limit larger than or equal to it.

   Since $\liminf s_n = \limsup s_n = s$, this implies that all subsequential limits must converge to $s$. No subsequence can converge to anything other than $s$, so the entire sequence must converge to $s$.

Thus, $(s_n)$ converges to $s$.

Problem 10:

Suppose that $x > 1$. Prove that $\lim_{n \to \infty} x^{1/n} = 1$.

Solution:

We need to prove that as $n \to \infty$, $x^{1/n}$ approaches 1 for any $x > 1$.

1. Rewrite $x^{1/n}$ as $\exp\left(\frac{\ln(x)}{n}\right)$.

2. As $n \to \infty$, the term $\frac{\ln(x)}{n}$ approaches 0 because $\ln(x)$ is a constant and $n$ grows indefinitely.

3. Therefore, $\exp\left(\frac{\ln(x)}{n}\right)$ approaches $\exp(0) = 1$.

Thus, we have $\lim_{n \to \infty} x^{1/n} = 1$.

Do you need further details or clarifications on any of these problems?

Follow-up Questions:

1. What is the formal definition of a subsequence?
2. Can you give another example where two subsequences are not equal?
3. How does the definition of $\liminf$ and $\limsup$ help in proving convergence?
4. What is the intuition behind the logarithmic approach in Problem 10?
5. How does boundedness of a sequence guarantee the existence of subsequential limits?

Tip:

Always check if a counterexample exists when a statement seems overly general, as it can save time compared to attempting a direct proof.