The image contains the statement and proof of Theorem 3.2.11 regarding sequences of positive real numbers and their limits. The theorem discusses the behavior of a sequence $(x_n)$ when the limit of the ratio $\lim \frac{x_{n+1}}{x_n} = L$ exists, particularly when $L < 1$.

Theorem 3.2.11:

Let $(x_n)$ be a sequence of positive real numbers such that $L := \lim \frac{x_{n+1}}{x_n}$ exists. If $L < 1$, then $(x_n)$ converges, and $\lim x_n = 0$.

Proof Sketch:

• From Theorem 3.2.4, it follows that $L \geq 0$.
• Select $r$ such that $L < r < 1$. Define $\epsilon := r - L > 0$.
• There exists a number $K \in \mathbb{N}$ such that for all $n \geq K$: $\left| \frac{x_{n+1}}{x_n} - L \right| < \epsilon$ This implies: $\frac{x_{n+1}}{x_n} < L + \epsilon = r$
• Therefore, for all $n \geq K$, we have: $0 < x_{n+1} < x_n r < x_{n-1} r^2 < \cdots < x_K r^{n-K+1}$
• Set $C := x_K r^{K}$. Then, for $n \geq K$, $x_n < Cr^{n+1}$. Since $0 < r < 1$, this implies $\lim x_n = 0$.

The proof concludes with the observation that the sequence tends to 0, using a comparison test and the fact that $r^n$ tends to 0 as $n \to \infty$.

Would you like any details on specific parts of the proof or the theorem?

Here are 5 related questions to deepen your understanding:

1. What does Theorem 3.2.4 state, and how is it used in this proof?
2. Why is it necessary to choose $r$ such that $L < r < 1$?
3. How does the inequality $\frac{x_{n+1}}{x_n} < r$ help in proving convergence of the sequence?
4. What role does the comparison sequence $x_n < Cr^{n+1}$ play in concluding $\lim x_n = 0$?
5. Can this theorem be extended to sequences with $L = 1$? What happens in that case?

Tip: When dealing with sequences and their limits, try to find a comparison sequence that is easier to analyze, as it often simplifies the proof of convergence or divergence.