Proving An Inequality Involving Lim Sup And Sequence Growth

Introduction

In real analysis, the study of sequences and their properties is a fundamental aspect of understanding mathematical concepts. One of the key tools used in this study is the concept of lim sup, which represents the limit superior of a sequence. In this article, we will explore an inequality involving lim sup and sequence growth, and provide a proof for this inequality.

The Inequality

The inequality we want to prove is given by:

$$(\forall n \in \mathbb{N}, a_n > 0) \implies \bigg(\limsup_{n \to \infty} a_n^{1/n} \leq \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}\bigg)$$

This inequality relates the growth rate of a sequence to the limit superior of the sequence. We will use this inequality to understand the relationship between the growth rate of a sequence and its limit superior.

Understanding Lim Sup

Before we dive into the proof, let's take a moment to understand what lim sup is. The limit superior of a sequence is the largest possible limit of the sequence. In other words, it is the largest value that the sequence can approach as it goes to infinity. The lim sup of a sequence is denoted by $\limsup_{n \to \infty} a_n$.

Proof of the Inequality

To prove the inequality, we will use a proof by contradiction. We will assume that the inequality is false, and then show that this assumption leads to a contradiction.

Assumption

Let's assume that the inequality is false. This means that there exists a sequence $(a_n)$ such that:

$$\limsup_{n \to \infty} a_n^{1/n} > \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$$

Construction of a New Sequence

We will construct a new sequence $(b_n)$ as follows:

$$b_n = \frac{a_{n+1}}{a_n}$$

This sequence is well-defined since $a_n > 0$ for all $n$. We will show that this sequence has a lim sup that is equal to the lim sup of the original sequence.

Properties of the New Sequence

The new sequence $(b_n)$ has the following properties:

• $b_n > 0$ for all $n$
• $\limsup_{n \to \infty} b_n = \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$

Contradiction

We will now show that the assumption that the inequality is false leads to a contradiction. Since $\limsup_{n \to \infty} a_n^{1/n} > \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$, we have:

$$\limsup_{n \to \infty} a_n^{1/n} > \limsup_{n \to \infty} b_n$$

However, since $b_n = \frac{a_{n+1}}{a_n}$, we have:

$$\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} b_n$$

This is a contradiction, since we have shown that $\limsup_{n \to \infty} a_n^{1/n} > \limsup_{n \to \infty} b_n$ and also that $\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} b_n$.

Conclusion

We have shown that the assumption that the inequality is false leads to a contradiction. Therefore, we can conclude that the inequality is true.

Implications of the Inequality

The inequality has several implications for the study of sequences and their properties. For example, it shows that the growth rate of a sequence is related to the limit superior of the sequence. This has important implications for the study of convergence and divergence of sequences.

Examples

To illustrate the inequality, let's consider a few examples.

Example 1

Let $(a_n)$ be the sequence defined by $a_n = n^2$. Then:

$$\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} n = \infty$$

and

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \limsup_{n \to \infty} \frac{(n+1)^2}{n^2} = \limsup_{n \to \infty} \frac{n^2+2n+1}{n^2} = 1$$

In this case, the inequality is satisfied.

Example 2

Let $(a_n)$ be the sequence defined by $a_n = 2^n$. Then:

$$\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} 2 = 2$$

and

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \limsup_{n \to \infty} \frac{2^{n+1}}{2^n} = \limsup_{n \to \infty} 2 = 2$$

In this case, the inequality is also satisfied.

Example 3

Let $(a_n)$ be the sequence defined by $a_n = (-1)^n$. Then:

$$\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} (-1)^{1/n} = -1$$

and

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \limsup_{n \to \infty} \frac{(-1)^{n+1}}{(-1)^n} = \limsup_{n \to \infty} (-1) = -1$$

In this case, the inequality is also satisfied.

Conclusion

Introduction

In our previous article, we proved an inequality involving lim sup and sequence growth. In this article, we will answer some common questions related to this inequality and provide additional insights into its proof.

Q: What is the significance of the inequality?

A: The inequality has important implications for the study of sequences and their properties. It shows that the growth rate of a sequence is related to the limit superior of the sequence, which is a fundamental concept in real analysis.

Q: Can you provide more examples to illustrate the inequality?

A: Yes, here are a few more examples:

Example 4

Let $(a_n)$ be the sequence defined by $a_n = n^3$. Then:

$$\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} n = \infty$$

and

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \limsup_{n \to \infty} \frac{(n+1)^3}{n^3} = \limsup_{n \to \infty} \frac{n^3+3n^2+3n+1}{n^3} = 1$$

In this case, the inequality is satisfied.

Example 5

Let $(a_n)$ be the sequence defined by $a_n = (-1)^n \cdot n^2$. Then:

$$\limsup_{n \to \infty} a_n^{1/n} = \limsup_{n \to \infty} (-1)^{1/n} \cdot n = -\infty$$

and

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \limsup_{n \to \infty} \frac{(-1)^{n+1} \cdot (n+1)^2}{(-1)^n \cdot n^2} = \limsup_{n \to \infty} \frac{(n+1)^2}{n^2} = 1$$

In this case, the inequality is also satisfied.

Q: Can you provide a proof of the inequality for the case where the sequence is bounded?

A: Yes, here is a proof of the inequality for the case where the sequence is bounded:

Let $(a_n)$ be a bounded sequence. Then, there exists a constant $M$ such that $|a_n| \leq M$ for all $n$. We can assume without loss of generality that $a_n > 0$ for all $n$.

We will show that the inequality is satisfied by considering the following two cases:

• Case 1: $\limsup_{n \to \infty} a_n^{1/n} = \infty$
• Case 2: $\limsup_{n \to \infty} a_n^{1/n} < \infty$

Case 1: $\limsup_{n \to \infty} a_n^{1/n} = \infty$

In this case, we have:

$$\limsup_{n \to \infty} a_n^{1/n} = \infty$$

Since the sequence is bounded, we have:

$$|a_n| \leq M$$

for all $n$. We can assume without loss of generality that $a_n > 0$ for all $n$. Then, we have:

$$a_n^{1/n} \geq \frac{a_n}{M}$$

for all $n$. Taking the limit superior of both sides, we get:

$$\limsup_{n \to \infty} a_n^{1/n} \geq \limsup_{n \to \infty} \frac{a_n}{M}$$

Since $\limsup_{n \to \infty} a_n^{1/n} = \infty$, we have:

$$\limsup_{n \to \infty} \frac{a_n}{M} = \infty$$

This implies that:

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \infty$$

Since $\limsup_{n \to \infty} a_n^{1/n} = \infty$, we have:

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} \geq \limsup_{n \to \infty} a_n^{1/n}$$

This shows that the inequality is satisfied in this case.

Case 2: $\limsup_{n \to \infty} a_n^{1/n} < \infty$

In this case, we have:

$$\limsup_{n \to \infty} a_n^{1/n} < \infty$$

Since the sequence is bounded, we have:

$$|a_n| \leq M$$

for all $n$. We can assume without loss of generality that $a_n > 0$ for all $n$. Then, we have:

$$a_n^{1/n} \leq \frac{a_n}{M}$$

for all $n$. Taking the limit superior of both sides, we get:

$$\limsup_{n \to \infty} a_n^{1/n} \leq \limsup_{n \to \infty} \frac{a_n}{M}$$

Since $\limsup_{n \to \infty} a_n^{1/n} < \infty$, we have:

$$\limsup_{n \to \infty} \frac{a_n}{M} < \infty$$

This implies that:

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} < \infty$$

Since $\limsup_{n \to \infty} a_n^{1/n} < \infty$, we have:

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} \leq \limsup_{n \to \infty} a_n^{1/n}$$

This shows that the inequality is satisfied in this case.

Q: Can you provide a proof of the inequality for the case where the sequence is unbounded?

A: Yes, here is a proof of the inequality for the case where the sequence is unbounded:

Let $(a_n)$ be an unbounded sequence. Then, there exists a subsequence $(a_{n_k})$ such that:

$$\lim_{k \to \infty} a_{n_k} = \infty$$

We can assume without loss of generality that $a_n > 0$ for all $n$.

We will show that the inequality is satisfied by considering the following two cases:

• Case 1: $\limsup_{n \to \infty} a_n^{1/n} = \infty$
• Case 2: $\limsup_{n \to \infty} a_n^{1/n} < \infty$

Case 1: $\limsup_{n \to \infty} a_n^{1/n} = \infty$

In this case, we have:

$$\limsup_{n \to \infty} a_n^{1/n} = \infty$$

Since the sequence is unbounded, we have:

$$\lim_{k \to \infty} a_{n_k} = \infty$$

We can assume without loss of generality that $a_{n_k} > 0$ for all $k$. Then, we have:

$$a_{n_k}^{1/n_k} \geq \frac{a_{n_k}}{M}$$

for all $k$, where $M$ is a constant such that $|a_n| \leq M$ for all $n$. Taking the limit superior of both sides, we get:

$$\limsup_{k \to \infty} a_{n_k}^{1/n_k} \geq \limsup_{k \to \infty} \frac{a_{n_k}}{M}$$

Since $\limsup_{k \to \infty} a_{n_k}^{1/n_k} = \infty$, we have:

$$\limsup_{k \to \infty} \frac{a_{n_k}}{M} = \infty$$

This implies that:

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = \infty$$

Since $\limsup_{n \to \infty} a_n^{1/n} = \infty$, we have:

$$\limsup_{n \to \infty} \frac{a_{n+1}}{a_n} \geq \limsup_{n \to \infty} a_n^{1/n}$$

This shows that the inequality is satisfied in this case.

Case 2: $\limsup_{n \to \infty} a_n^{1/n} < \infty$

In this case, we have:

$$\limsup_{n \to \infty} a_n^{1/n} < \infty$$

Since the sequence is un