Prove Sequence Convergence Using Liminf Limsup And Cauchy Arguments

#Introduction In the realm of mathematical analysis, determining the convergence of a sequence is a fundamental task. Several tools and techniques are available, each with its strengths and applicability. Among these, the concepts of limit inferior (liminf) and limit superior (limsup) offer powerful ways to analyze the long-term behavior of sequences, especially those that might not have a traditional limit. Another crucial concept is that of a Cauchy sequence, which provides an intrinsic criterion for convergence without needing to know the limit beforehand. This article delves into a specific problem: converting a liminf = limsup argument into a Cauchy argument to prove the convergence of a sequence. This approach highlights the interconnectedness of these concepts and offers a deeper understanding of sequence convergence.

We will explore the problem where a sequence $(x_n)_{n=1}^\infty$ satisfies the condition $0 \leq x_{n+1} \leq x_n + \frac{1}{n^2}$ for all $n \in \mathbb{N}$. The goal is to demonstrate that $\displaystyle\lim_{n\to\infty} x_n$ exists by leveraging both the liminf = limsup approach and the Cauchy sequence criterion. This exploration will not only provide a solution to the problem but also illuminate the general strategies for tackling similar convergence problems.

#Problem Statement Let's formally state the problem we aim to solve. Consider a sequence $(x_n)_{n=1}^\infty$ of real numbers such that:

$0 \leq x_{n+1} \leq x_n + \frac{1}{n^2} \quad \forall n \in \mathbb{N}$

Our objective is to prove that the sequence $(x_n)$ converges, that is, $\displaystyle\lim_{n\to\infty} x_n$ exists. We will approach this problem using two main strategies:

1. Demonstrating that liminf and limsup of the sequence are equal.
2. Showing that the sequence is a Cauchy sequence.

The equivalence between a convergent sequence and a Cauchy sequence is a cornerstone of real analysis. A sequence is Cauchy if its terms become arbitrarily close to each other as the index increases. This property is particularly useful because it allows us to establish convergence without explicitly knowing the limit. The liminf and limsup provide information about the eventual range of the sequence. If they coincide, it strongly suggests that the sequence converges to that common value. The challenge lies in effectively using the given inequality to bridge these concepts and arrive at a rigorous proof of convergence. We will meticulously construct the arguments, highlighting the key steps and reasoning involved in each approach.

To demonstrate convergence using the liminf = limsup argument, we aim to show that the limit inferior and limit superior of the sequence $(x_n)$ are equal. This equality implies that the sequence converges to a unique limit. Let's first define these concepts formally.

Definition of Limit Inferior (liminf) and Limit Superior (limsup) The limit superior of a sequence $(x_n)$, denoted as $\limsup_{n\to\infty} x_n$, is the largest limit of any convergent subsequence of $(x_n)$. Formally,

$\limsup_{n\to\infty} x_n = \lim_{n\to\infty} (\sup_{k\geq n} x_k)$

Similarly, the limit inferior of a sequence $(x_n)$, denoted as $\liminf_{n\to\infty} x_n$, is the smallest limit of any convergent subsequence of $(x_n)$. Formally,

$\liminf_{n\to\infty} x_n = \lim_{n\to\infty} (\inf_{k\geq n} x_k)$

If $\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n = L$, then the sequence $(x_n)$ converges to $L$.

Applying the Definition to Our Sequence Given the inequality $0 \leq x_{n+1} \leq x_n + \frac{1}{n^2}$, we can analyze the behavior of the sequence. First, observe that the sequence $(x_n)$ is bounded below by 0. Next, we need to investigate how the terms of the sequence evolve as $n$ increases. We can rewrite the given inequality to understand the cumulative effect of the $\frac{1}{n^2}$ terms.

For any $m > n$, we can iterate the inequality:

$x_{n+1} \leq x_n + \frac{1}{n^2}$

$x_{n+2} \leq x_{n+1} + \frac{1}{(n+1)^2} \leq x_n + \frac{1}{n^2} + \frac{1}{(n+1)^2}$

Continuing this process, we get:

$x_m \leq x_n + \sum_{k=n}^{m-1} \frac{1}{k^2}$

Now, let $m$ approach infinity. The series $\sum_{k=1}^\infty \frac{1}{k^2}$ is a well-known convergent series (it converges to $\frac{\pi^2}{6}$). Therefore, the tail of the series, $\sum_{k=n}^\infty \frac{1}{k^2}$, converges to 0 as $n$ goes to infinity. This fact is crucial because it allows us to control the difference between terms in the sequence.

Taking the limit as $m \to \infty$, we obtain:

$\limsup_{m\to\infty} x_m \leq x_n + \sum_{k=n}^\infty \frac{1}{k^2}$

Now, let's consider $\limsup_{n\to\infty} x_n$. We know that the sequence $(x_n)$ is bounded below, so $\limsup_{n\to\infty} x_n$ exists. Taking the limit as $n \to \infty$ on both sides, we get:

$\limsup_{n\to\infty} x_n \leq \liminf_{n\to\infty} x_n + \lim_{n\to\infty} \sum_{k=n}^\infty \frac{1}{k^2}$

Since $\lim_{n\to\infty} \sum_{k=n}^\infty \frac{1}{k^2} = 0$, we have:

$\limsup_{n\to\infty} x_n \leq \liminf_{n\to\infty} x_n$

However, we also know that, by definition, $\liminf_{n\to\infty} x_n \leq \limsup_{n\to\infty} x_n$. Combining these two inequalities, we conclude that:

$\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n$

This equality implies that the sequence $(x_n)$ converges. This method elegantly uses the properties of limsup and liminf along with the convergence of the series $\sum_{k=1}^\infty \frac{1}{k^2}$ to establish the convergence of the sequence.

An alternative approach to proving the convergence of the sequence $(x_n)$ is to demonstrate that it is a Cauchy sequence. This method relies on the completeness of the real numbers, which guarantees that every Cauchy sequence in $\mathbb{R}$ converges. Let's first define what a Cauchy sequence is.

Definition of a Cauchy Sequence A sequence $(x_n)$ is said to be a Cauchy sequence if for every $\epsilon > 0$, there exists a positive integer $N$ such that for all $m, n > N$, we have:

$|x_m - x_n| < \epsilon$

In essence, this definition means that the terms of the sequence become arbitrarily close to each other as the indices increase. To prove that our sequence $(x_n)$ is Cauchy, we need to show that it satisfies this condition.

Applying the Cauchy Criterion to Our Sequence Recall the given inequality: $0 \leq x_{n+1} \leq x_n + \frac{1}{n^2}$. Without loss of generality, assume $m > n$. From our previous analysis, we have derived the inequality:

$x_m \leq x_n + \sum_{k=n}^{m-1} \frac{1}{k^2}$

However, we also need a lower bound for $x_m$. Since $x_{n+1} \geq 0$, we know that the sequence is bounded below by 0, but we need to relate $x_m$ and $x_n$ more directly. By reversing the iterative process and noting that the terms are non-negative, we can establish a lower bound.

Consider $m > n$. We know $x_m \geq 0$. We want to find a lower bound in terms of $x_n$. However, the given inequality directly provides an upper bound. To find a lower bound, we can use the fact that the sequence terms are non-negative, but this doesn't directly relate $x_m$ to $x_n$. Instead, let's focus on estimating $|x_m - x_n|$.

From the inequality $x_m \leq x_n + \sum_{k=n}^{m-1} \frac{1}{k^2}$, we can write:

$x_m - x_n \leq \sum_{k=n}^{m-1} \frac{1}{k^2}$

Since $x_{n+1} \leq x_n + \frac{1}{n^2}$, we also have $x_n - x_m \leq \sum_{k=n}^{m-1} \frac{1}{k^2}$ is not necessarily true since we don't have information on how much $x_n$ decreases. But, since $0 \leq x_{n+1}$, the sequence $x_n$ is bounded below by 0.

Therefore, we consider $|x_m - x_n|$. We have:

$|x_m - x_n| \leq \sum_{k=n}^{m-1} \frac{1}{k^2}$

Now, we need to show that this sum can be made arbitrarily small for sufficiently large $n$. Recall that the series $\sum_{k=1}^\infty \frac{1}{k^2}$ converges. Therefore, for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n > N$,

$\sum_{k=n}^\infty \frac{1}{k^2} < \epsilon$

Since $\sum_{k=n}^{m-1} \frac{1}{k^2}$ is a partial sum of the convergent series, we have for $m > n > N$:

$\sum_{k=n}^{m-1} \frac{1}{k^2} < \sum_{k=n}^\infty \frac{1}{k^2} < \epsilon$

Thus, for all $m, n > N$,

$|x_m - x_n| \leq \sum_{k=n}^{m-1} \frac{1}{k^2} < \epsilon$

This demonstrates that the sequence $(x_n)$ satisfies the Cauchy criterion. Therefore, by the completeness of the real numbers, the sequence $(x_n)$ converges. This method directly uses the given inequality and the properties of convergent series to establish the Cauchy property, providing a robust proof of convergence.

In this article, we explored two distinct methods to prove the convergence of a sequence $(x_n)$ satisfying the condition $0 \leq x_{n+1} \leq x_n + \frac{1}{n^2}$ for all $n \in \mathbb{N}$. The first approach leveraged the concepts of limit inferior (liminf) and limit superior (limsup), demonstrating that their equality implies convergence. By carefully manipulating the given inequality and using the convergence of the series $\sum_{k=1}^\infty \frac{1}{k^2}$, we showed that $\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n$, thus establishing the convergence of $(x_n)$.

The second method employed the Cauchy sequence criterion. This approach, rooted in the completeness of the real numbers, required us to show that the terms of the sequence become arbitrarily close to each other as the indices increase. By again utilizing the given inequality and the convergence of the series $\sum_{k=1}^\infty \frac{1}{k^2}$, we demonstrated that $(x_n)$ is indeed a Cauchy sequence, thereby proving its convergence.

Both methods highlight the interconnectedness of different concepts in real analysis. The liminf = limsup argument provides a powerful tool for analyzing the long-term behavior of sequences, while the Cauchy sequence criterion offers an intrinsic condition for convergence. The ability to convert between these arguments not only deepens our understanding of sequence convergence but also equips us with a versatile toolkit for tackling a wide range of problems in mathematical analysis.

This exploration underscores the importance of having multiple perspectives when approaching mathematical problems. While one method might be more direct or intuitive in a given situation, understanding alternative approaches enriches our mathematical understanding and problem-solving capabilities. The specific problem we addressed serves as a valuable case study for applying these concepts and techniques in practice, fostering a deeper appreciation for the intricacies of sequence convergence.