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, where $\{y_n\}$ 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 begin the proof, let's recall some definitions and results that we will need.

• A sequence $\{y_n\}$ is said to be strictly monotonically increasing if $y_n < y_{n+1}$ for all $n \in \mathbb{N}$.

• The limit of a sequence $\{x_n\}$ is denoted by $\lim_{n \to \infty} x_n = L$ if for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $|x_n - L| < \epsilon$ for all $n \geq N$.

• The upper and lower limits of a sequence $\{x_n\}$ are denoted by $\limsup_{n \to \infty} x_n$ and $\liminf_{n \to \infty} x_n$, respectively. They are defined as follows:

  • $\limsup_{n \to \infty} x_n = \inf_{n \geq 1} \sup_{k \geq n} x_k$
  • $\liminf_{n \to \infty} x_n = \sup_{n \geq 1} \inf_{k \geq n} x_k$

Proof of the Stolz Theorem

Theorem

Let $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ be two real sequences, and let $\alpha \in \mathbb{R}$. If $\{y_n\}$ is strictly monotonically increasing, then $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$ if and only if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.

Proof

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 $|\frac{x_n}{y_n} - \alpha| < \epsilon$ for all $n \geq N$. This implies that $|\frac{x_n}{y_n} - \alpha| < \epsilon$ for all $n \geq N$.

Now, let $n \geq N$ be given. We have:

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

Since $\{y_n\}$ is strictly monotonically increasing, we have $y_n < y_{n+1}$ for all $n \in \mathbb{N}$. Therefore, we have:

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

Since $|\frac{x_n}{y_n} - \alpha| < \epsilon$ for all $n \geq N$, we have:

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

Since $\{y_n\}$ is strictly monotonically increasing, we have $y_n < y_{n+1}$ for all $n \in \mathbb{N}$. Therefore, we have:

$$\left|\frac{y_n - y_{n+1}}{y_{n+1} - y_n}\right| = \frac{y_n - y_{n+1}}{y_{n+1} - y_n} = -1$$

Therefore, we have:

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

Since $\epsilon > 0$ is arbitrary, we have:

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

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 $|\frac{x_{n+1} - x_n}{y_{n+1} - y_n} - \alpha| < \epsilon$ for all $n \geq N$. This implies that $|\frac{x_{n+1} - x_n}{y_{n+1} - y_n} - \alpha| < \epsilon$ for all $n \geq N$.

Now, let $n \geq N$ be given. We have:

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

Since $|\frac{x_{n+1} - x_n}{y_{n+1} - y_n} - \alpha| < \epsilon$ for all $n \geq N$, we have:

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

Since $|\frac{x_{n+1} - x_n}{y_{n+1} - y_n} - \alpha| < \epsilon$ for all $n \geq N$, we have:

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

Since $|\frac{x_{n+1} - x_n}{y_{n+1} - y_n} - \alpha| < \epsilon$ for all $n \geq N$, we have:

$$\left|\frac{x_{n-1}}{y_{n-1}} - \alpha\right| < \epsilon + \left|\frac{x_{n-2}}{y_{n-2}} - \alpha\right|$$

Continuing in this manner, we have:

$$\left|\frac{x_n}{y_n} - \alpha\right| < \epsilon + \epsilon + \epsilon + \cdots + \epsilon + \left|\frac{x_1}{y_1} - \alpha\right|$$

Since $\epsilon > 0$ is arbitrary, we have:

$$\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$$

Conclusion

Introduction

The Stolz theorem is a fundamental result in real analysis that provides a criterion for the convergence of a sequence of real numbers. In our previous article, we proved the Stolz theorem using the upper and lower limits. In this article, we will answer some frequently asked questions about the Stolz theorem.

Q: What is the Stolz theorem?

A: The Stolz theorem is a 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, where $\{y_n\}$ 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 is the significance of the Stolz theorem?

A: The Stolz theorem is significant because it provides a criterion for the convergence of a sequence of real numbers. It is a powerful tool in real analysis and has many applications in mathematics and physics.

Q: What are the assumptions of the Stolz theorem?

A: The assumptions of the Stolz theorem are:

• $\{x_n\}$ and $\{y_n\}$ are two real sequences.
• $\{y_n\}$ is strictly monotonically increasing.
• $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.

Q: How do I apply the Stolz theorem?

A: To apply the Stolz theorem, you need to:

1. Check if the sequence $\{y_n\}$ is strictly monotonically increasing.
2. Check if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.
3. If both conditions are satisfied, then $\lim_{n \to \infty} \frac{x_n}{y_n} = \alpha$.

Q: What are some common mistakes to avoid when applying the Stolz theorem?

A: Some common mistakes to avoid when applying the Stolz theorem are:

• Not checking if the sequence $\{y_n\}$ is strictly monotonically increasing.
• Not checking if $\lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n} = \alpha$.
• Not using the correct definition of the Stolz theorem.

Q: Can the Stolz theorem be used to prove other theorems?

A: Yes, the Stolz theorem can be used to prove other theorems. For example, it can be used to prove the Cauchy criterion for convergence.

Q: What are some applications of the Stolz theorem?

A: The Stolz theorem has many applications in mathematics and physics. Some examples include:

• Convergence of sequences and series.
• Limit theorems.
• Approximation of functions.

Conclusion

In this article, we have answered some frequently asked questions about the Stolz theorem. The Stolz theorem is a powerful tool in real analysis that provides a criterion for the convergence of a sequence of real numbers. It has many applications in mathematics and physics and is a fundamental result in real analysis.

Further Reading

For further reading on the Stolz theorem, we recommend the following resources:

• "Real Analysis" by Walter Rudin.
• "Introduction to Real Analysis" by Bartle and Sherbert.
• "Real Analysis: A First Course" by Richard R. Goldberg.

We hope this article has been helpful in understanding the Stolz theorem. If you have any further questions, please don't hesitate to ask.