Prove The Stolz Theorem Using The Upper And Lower Limits.

Introduction

The Stolz theorem is a fundamental result in real analysis that provides a criterion for the convergence of a sequence of real numbers. It states that if $\{x_n\}$ and $\{y_n\}$ are two real sequences, and $\{y_n\}$ is strictly monotonically increasing, then $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ if and only if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$. In this article, we will prove the Stolz theorem using the upper and lower limits.

Preliminaries

Before we proceed with the proof, let us recall some basic definitions and results from real analysis.

• A sequence $\{x_n\}$ is said to be strictly monotonically increasing if $x_n < x_{n+1}$ for all $n \in \mathbb{N}$.
• The limit superior of a sequence $\{x_n\}$, denoted by $\limsup_{n \to \infty} x_n$, is the greatest limit point of the sequence.
• The limit inferior of a sequence $\{x_n\}$, denoted by $\liminf_{n \to \infty} x_n$, is the smallest limit point of the sequence.
• A sequence $\{x_n\}$ is said to be convergent if there exists a real number $x$ such that $\lim_{n \to \infty} x_n = x$.

Proof of the Stolz Theorem

Necessity

Suppose that $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$. We need to show that $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.

Let $\epsilon > 0$ be given. Since $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, we have

$$\left| \frac{x_n}{y_n} - \alpha \right| < \epsilon.$$

Now, for any $n \geq N$, we have

$$\frac{x_n}{y_n} - \alpha = \frac{x_n - \alpha y_n}{y_n} = \frac{x_n - \alpha y_n + \alpha y_n - \alpha y_{n-1}}{y_n} - \frac{\alpha y_n - \alpha y_{n-1}}{y_n}.$$

Using the triangle inequality, we get

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \left| \frac{x_n - \alpha y_n + \alpha y_n - \alpha y_{n-1}}{y_n} \right| + \left| \frac{\alpha y_n - \alpha y_{n-1}}{y_n} \right|.$$

Simplifying, we get

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \left| \frac{x_n - \alpha y_n}{y_n} \right| + \left| \frac{\alpha (y_n - y_{n-1})}{y_n} \right|.$$

Now, since $\lim_{n \to \infty} y_n = +\infty$, there exists $M \in \mathbb{N}$ such that for all $n \geq M$, we have $y_n > \frac{1}{\epsilon}$. Therefore, for all $n \geq \max\{N, M\}$, we have

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \left| \frac{x_n - \alpha y_n}{y_n} \right| + \epsilon \left| \frac{y_n - y_{n-1}}{y_n} \right|.$$

Now, since $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$, we have

$$\lim_{n \to \infty} \left| \frac{x_n - \alpha y_n}{y_n} \right| = 0.$$

Therefore, for all $n \geq \max\{N, M\}$, we have

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \epsilon \left| \frac{y_n - y_{n-1}}{y_n} \right|.$$

Now, since $\lim_{n \to \infty} y_n = +\infty$, we have

$$\lim_{n \to \infty} \left| \frac{y_n - y_{n-1}}{y_n} \right| = 0.$$

Therefore, for all $n \geq \max\{N, M\}$, we have

$$\left| \frac{x_n}{y_n} - \alpha \right| < \epsilon.$$

This shows that $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.

Sufficiency

Suppose that $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$. We need to show that $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$.

Let $\epsilon > 0$ be given. Since $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, we have

$$\left| \frac{x_{n+1} - x_n}{y_{n+1} - y_n} - \alpha \right| < \epsilon.$$

Now, for any $n \geq N$, we have

$$\frac{x_n}{y_n} - \alpha = \frac{x_n - \alpha y_n}{y_n} = \frac{x_n - \alpha y_n + \alpha y_n - \alpha y_{n-1}}{y_n} - \frac{\alpha y_n - \alpha y_{n-1}}{y_n}.$$

Using the triangle inequality, we get

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \left| \frac{x_n - \alpha y_n + \alpha y_n - \alpha y_{n-1}}{y_n} \right| + \left| \frac{\alpha y_n - \alpha y_{n-1}}{y_n} \right|.$$

Simplifying, we get

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \left| \frac{x_n - \alpha y_n}{y_n} \right| + \left| \frac{\alpha (y_n - y_{n-1})}{y_n} \right|.$$

Now, since $\lim_{n \to \infty} y_n = +\infty$, there exists $M \in \mathbb{N}$ such that for all $n \geq M$, we have $y_n > \frac{1}{\epsilon}$. Therefore, for all $n \geq \max\{N, M\}$, we have

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \left| \frac{x_n - \alpha y_n}{y_n} \right| + \epsilon \left| \frac{y_n - y_{n-1}}{y_n} \right|.$$

Now, since $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$, we have

$$\lim_{n \to \infty} \left| \frac{x_n - \alpha y_n}{y_n} \right| = 0.$$

Therefore, for all $n \geq \max\{N, M\}$, we have

$$\left| \frac{x_n}{y_n} - \alpha \right| \leq \epsilon \left| \frac{y_n - y_{n-1}}{y_n} \right|.$$

Now, since $\lim_{n \to \infty} y_n = +\infty$, we have

$$\lim_{n \to \infty} \left| \frac{y_n - y_{n-1}}{y_n} \right| = 0.$$

Therefore, for all $n \geq \max\{N, M\}$, we have

\left| \frac{x_n}{y_n}<br/> **Q&A: Proving the Stolz Theorem** =====================================

Q: What is the Stolz Theorem?

A: The Stolz Theorem is a fundamental result in real analysis that provides a criterion for the convergence of a sequence of real numbers. It states that if $\{x_n\}$ and $\{y_n\}$ are two real sequences, and $\{y_n\}$ is strictly monotonically increasing, then $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ if and only if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.

Q: What are the conditions for the Stolz Theorem to hold?

A: The Stolz Theorem holds if and only if the following conditions are met:

• $\{y_n\}$ is strictly monotonically increasing.
• $\lim_{n \to \infty} y_n = +\infty$.

Q: How do I prove the Stolz Theorem?

A: To prove the Stolz Theorem, you need to show that $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ if and only if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$. This can be done by using the definition of limit and the properties of sequences.

Q: What are the key steps in proving the Stolz Theorem?

A: The key steps in proving the Stolz Theorem are:

• Showing that $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ implies $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.
• Showing that $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$ implies $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$.

Q: How do I use the Stolz Theorem in practice?

A: The Stolz Theorem can be used to prove the convergence of a sequence of real numbers. For example, if you have a sequence $\{x_n\}$ and a sequence $\{y_n\}$ that is strictly monotonically increasing, you can use the Stolz Theorem to show that $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ if and only if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.

Q: What are some common applications of the Stolz Theorem?

A: The Stolz Theorem has many applications in real analysis, including:

• Proving the convergence of sequences of real numbers.
• Proving the convergence of series of real numbers.
• Studying the properties of limit points of sequences.

Q: How do I find the limit of a sequence using the Stolz Theorem?

A: To find the limit of a sequence using the Stolz Theorem, you need to follow these steps:

1. Check if the sequence $\{y_n\}$ is strictly monotonically increasing.
2. Check if $\lim_{n \to \infty} y_n = +\infty$.
3. If both conditions are met, use the Stolz Theorem to show that $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ if and only if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.
4. Use the definition of limit to find the value of $\alpha$.

Q: What are some common mistakes to avoid when using the Stolz Theorem?

A: Some common mistakes to avoid when using the Stolz Theorem include:

• Assuming that the sequence $\{y_n\}$ is strictly monotonically increasing without checking.
• Assuming that $\lim_{n \to \infty} y_n = +\infty$ without checking.
• Not using the definition of limit to find the value of $\alpha$.

By following these steps and avoiding common mistakes, you can use the Stolz Theorem to prove the convergence of sequences of real numbers and find the limit of a sequence.