Proving Divergence Of Series A Comprehensive Guide

In the realm of mathematical analysis, determining the convergence or divergence of a sequence is a fundamental task. This article delves into the fascinating problem of proving the divergence of a recursively defined sequence. Specifically, we will explore a sequence presented in the 2024 UTCN competition, offering a detailed solution and insights into the techniques used. Our focus is on providing a clear and comprehensive understanding of the methods involved in proving divergence, making this article an invaluable resource for students, educators, and anyone with an interest in mathematical problem-solving.

Consider the sequence $(x_n)_{n \geq 1}$ defined recursively as follows:

$$x_1 = \sqrt{2}, \quad x_{n+1} = x_n + \frac{1}{\sqrt{n} \cdot x_n} \quad \text{for } n \ge 1$$

The core question we aim to address is: How can we rigorously prove that this sequence $(x_n)$ diverges? This problem is not just an academic exercise; it exemplifies the kind of challenges encountered in advanced mathematical competitions and research. The recursive nature of the sequence, combined with the presence of the square root and the reciprocal, adds layers of complexity that demand a careful and strategic approach. By tackling this problem, we will explore various techniques and principles that are essential for handling similar divergence proofs.

Initial Observations and Strategy

Before diving into a formal proof, let's make some initial observations. The recursive definition $x_{n+1} = x_n + \frac{1}{\sqrt{n} \cdot x_n}$ suggests that each term is obtained by adding a positive quantity to the previous term. This immediately tells us that the sequence is strictly increasing. However, simply being increasing does not guarantee divergence; a sequence could be increasing and still converge to a finite limit (e.g., $x_n = 1 - \frac{1}{n}$). To prove divergence, we need to show that the sequence does not have an upper bound or that it grows without bound.

A common strategy for proving divergence is to show that the terms of the sequence grow sufficiently fast. In this case, we will attempt to find a lower bound for $x_n$ that grows to infinity as $n$ goes to infinity. This will demonstrate that the sequence cannot converge to a finite limit and thus must diverge.

Rigorous Proof

To prove the divergence of the sequence $(x_n)$, we need to show that $\lim_{n \to \infty} x_n = \infty$. This means that for any arbitrarily large number $M$, we can find an index $N$ such that for all $n > N$, $x_n > M$. To achieve this, we'll derive a useful inequality that provides a lower bound for the growth of $x_n$.

Step 1: Squaring the Recursive Relation

Let's square both sides of the recursive relation to eliminate the square root in the denominator:

$$x_{n+1}^2 = \left(x_n + \frac{1}{\sqrt{n} \cdot x_n}\right)^2$$

Expanding the square, we get:

$$x_{n+1}^2 = x_n^2 + 2 \cdot x_n \cdot \frac{1}{\sqrt{n} \cdot x_n} + \frac{1}{n \cdot x_n^2}$$

Simplifying the expression:

$$x_{n+1}^2 = x_n^2 + \frac{2}{\sqrt{n}} + \frac{1}{n \cdot x_n^2}$$

Step 2: Establishing a Lower Bound

Since $x_n$ is strictly increasing and $x_1 = \sqrt{2}$, we know that $x_n \geq \sqrt{2}$ for all $n \geq 1$. Therefore, the term $\frac{1}{n \cdot x_n^2}$ is positive, and we can establish a lower bound by ignoring it:

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

This inequality is crucial because it relates the square of the $(n+1)$-th term to the square of the $n$-th term, providing a means to track the growth of the sequence.

Step 3: Summing the Inequality

Now, let's sum this inequality from $n = 1$ to $n = k - 1$ (where $k > 1$):

$$\sum_{n=1}^{k-1} x_{n+1}^2 > \sum_{n=1}^{k-1} \left(x_n^2 + \frac{2}{\sqrt{n}}\right)$$

This sum can be expanded as follows:

$$x_2^2 + x_3^2 + \cdots + x_k^2 > (x_1^2 + x_2^2 + \cdots + x_{k-1}^2) + 2 \sum_{n=1}^{k-1} \frac{1}{\sqrt{n}}$$

Notice that most of the terms cancel out, leaving us with:

$$x_k^2 > x_1^2 + 2 \sum_{n=1}^{k-1} \frac{1}{\sqrt{n}}$$

Since $x_1 = \sqrt{2}$, we have $x_1^2 = 2$. Thus:

$$x_k^2 > 2 + 2 \sum_{n=1}^{k-1} \frac{1}{\sqrt{n}}$$

Step 4: Estimating the Sum

We now need to estimate the sum $\sum_{n=1}^{k-1} \frac{1}{\sqrt{n}}$. This sum can be bounded below using an integral. Specifically, we can compare the sum to the integral of the function $f(x) = \frac{1}{\sqrt{x}}$:

$$\sum_{n=1}^{k-1} \frac{1}{\sqrt{n}} = 1 + \sum_{n=2}^{k-1} \frac{1}{\sqrt{n}} > \int_{1}^{k} \frac{1}{\sqrt{x}} dx$$

The integral can be computed as follows:

$$\int_{1}^{k} \frac{1}{\sqrt{x}} dx = \int_{1}^{k} x^{-\frac{1}{2}} dx = \left[2\sqrt{x}\right]_1^k = 2\sqrt{k} - 2$$

Therefore:

$$\sum_{n=1}^{k-1} \frac{1}{\sqrt{n}} > 2\sqrt{k} - 2$$

Step 5: Establishing the Divergence

Substituting this estimate back into our inequality, we get:

$$x_k^2 > 2 + 2(2\sqrt{k} - 2) = 4\sqrt{k} - 2$$

Taking the square root of both sides:

$$x_k > \sqrt{4\sqrt{k} - 2}$$

As $k$ approaches infinity, $\sqrt{4\sqrt{k} - 2}$ also approaches infinity. This means that for any large number $M$, we can find a $K$ such that for all $k > K$, $x_k > M$. Therefore, the sequence $(x_n)$ diverges.

Conclusion

We have successfully proven that the recursively defined sequence $(x_n)$ diverges by establishing a lower bound that grows without bound. The key steps in our proof included squaring the recursive relation, summing the resulting inequality, estimating the sum using an integral, and showing that the lower bound grows to infinity. This problem illustrates the power of analytical techniques in proving divergence and provides valuable insights into handling recursively defined sequences. Understanding these methods is essential for advanced problem-solving in mathematics and related fields.

While the approach outlined above provides a rigorous proof of divergence, there are alternative methods and further insights that can enhance our understanding of this problem and similar sequences. Let's explore some of these.

Stolz-Cesàro Theorem

One alternative method to prove the divergence of $(x_n)$ is by using the Stolz-Cesàro Theorem. This theorem is a powerful tool for evaluating limits of sequences and can be particularly useful when dealing with recursive definitions. The theorem states:

Theorem (Stolz-Cesàro): Let $(a_n)$ and $(b_n)$ be sequences of real numbers. If $(b_n)$ is strictly monotone and divergent, and the limit

$$\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L$$

exists (finite or infinite), then

$$\lim_{n \to \infty} \frac{a_n}{b_n} = L$$

To apply this theorem to our problem, we can consider the sequence $(x_n^2)$. We know that $(x_n)$ is strictly increasing and diverges, so $(x_n^2)$ also diverges. Let's analyze the limit of the difference quotient:

$$\lim_{n \to \infty} \frac{x_{n+1}^2 - x_n^2}{n+1 - n} = \lim_{n \to \infty} (x_{n+1}^2 - x_n^2)$$

Using the recursive relation, we have:

$$x_{n+1}^2 - x_n^2 = \left(x_n + \frac{1}{\sqrt{n} \cdot x_n}\right)^2 - x_n^2 = x_n^2 + \frac{2}{\sqrt{n}} + \frac{1}{n x_n^2} - x_n^2 = \frac{2}{\sqrt{n}} + \frac{1}{n x_n^2}$$

Now, we need to show that $\lim_{n \to \infty} (x_{n+1}^2 - x_n^2)$ exists. We already have the inequality:

$$x_k^2 > 4\sqrt{k} - 2$$

From this, we can infer that $x_n > \sqrt{4\sqrt{n} - 2}$ for large $n$. Thus, $x_n$ grows at least as fast as $n^{1/4}$. Therefore, the term $\frac{1}{n x_n^2}$ will approach 0 faster than $\frac{2}{\sqrt{n}}$ as $n \to \infty$.

So, we have:

$$\lim_{n \to \infty} (x_{n+1}^2 - x_n^2) = \lim_{n \to \infty} \left(\frac{2}{\sqrt{n}} + \frac{1}{n x_n^2}\right) = 0$$

However, this approach does not directly help us prove divergence. Instead, we can consider a different application of the Stolz-Cesàro theorem. Let's consider the limit:

$$\lim_{n \to \infty} \frac{x_n^2}{n}$$

Using the Stolz-Cesàro theorem with $a_n = x_n^2$ and $b_n = n$, we have:

$$\lim_{n \to \infty} \frac{x_{n+1}^2 - x_n^2}{n+1 - n} = \lim_{n \to \infty} \frac{2}{\sqrt{n}} + \frac{1}{n x_n^2}$$

From our previous analysis, we know that this limit is 0. Thus,

$$\lim_{n \to \infty} \frac{x_n^2}{n} = 0$$

This result tells us that $x_n$ grows slower than $\sqrt{n}$. However, it doesn't directly prove divergence. To prove divergence using this approach, we need a different comparison.

Let's consider the original inequality:

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

Divide both sides by $n$:

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

Summing this inequality from $n=1$ to $k-1$:

$$\sum_{n=1}^{k-1} \frac{x_{n+1}^2}{n} > \sum_{n=1}^{k-1} \left(\frac{x_n^2}{n} + \frac{2}{n\sqrt{n}}\right)$$

This approach becomes complex and doesn't lead to a straightforward proof of divergence in this manner. The Stolz-Cesàro theorem can provide insights into the growth rate of $x_n$, but it's not the most direct method for proving divergence in this case.

Asymptotic Analysis

Another perspective is to consider the asymptotic behavior of $x_n$. From the recursive relation, we have:

$$x_{n+1} = x_n + \frac{1}{\sqrt{n} x_n}$$

We've shown that $x_n$ diverges, so we can assume that $x_n$ grows without bound. Let's assume that $x_n \sim n^\alpha$ for some $\alpha > 0$. Then, we have:

$$x_{n+1} - x_n \approx (n+1)^\alpha - n^\alpha \approx \alpha n^{\alpha-1}$$

From the recursive relation, we also have:

$$x_{n+1} - x_n = \frac{1}{\sqrt{n} x_n} \approx \frac{1}{\sqrt{n} n^\alpha} = n^{-\frac{1}{2} - \alpha}$$

Equating the exponents, we get:

$$\alpha - 1 = -\frac{1}{2} - \alpha$$

$$2\alpha = \frac{1}{2}$$

$$\alpha = \frac{1}{4}$$

This suggests that $x_n$ grows approximately as $n^{1/4}$. To verify this, we can substitute $\alpha = \frac{1}{4}$ back into the recursive relation:

$$x_{n+1} - x_n \approx \frac{1}{4} n^{-\frac{3}{4}}$$

And:

$$\frac{1}{\sqrt{n} x_n} \approx \frac{1}{\sqrt{n} n^{\frac{1}{4}}} = n^{-\frac{3}{4}}$$

These approximations are consistent, which supports the idea that $x_n$ grows like $n^{1/4}$. While this asymptotic analysis provides valuable intuition about the growth of the sequence, it is not a rigorous proof of divergence on its own. It gives us a sense of how the sequence behaves for large $n$, which can guide us in constructing a formal proof.

Integral Comparison Test

The approach we used in the main proof, where we estimated the sum using an integral, is related to the Integral Comparison Test. This test is a powerful tool for determining the convergence or divergence of series. It states:

Theorem (Integral Comparison Test): Let $f$ be a continuous, positive, and decreasing function on the interval $[1, \infty)$. Let $a_n = f(n)$ for all positive integers $n$. Then, the series $\sum_{n=1}^{\infty} a_n$ converges if and only if the integral $\int_{1}^{\infty} f(x) dx$ converges.

In our case, we considered the sum $\sum_{n=1}^{k-1} \frac{1}{\sqrt{n}}$, which can be compared to the integral $\int_{1}^{k} \frac{1}{\sqrt{x}} dx$. Since this integral diverges as $k \to \infty$, the sum also diverges. This is a key step in our proof and highlights the utility of the Integral Comparison Test in analyzing the behavior of sums and sequences.

More Refined Lower Bounds

We established a lower bound for $x_n$ by summing the inequality $x_{n+1}^2 > x_n^2 + \frac{2}{\sqrt{n}}$. It may be possible to find a more refined lower bound by considering the term $\frac{1}{n x_n^2}$ that we initially ignored. However, this often leads to more complex calculations without significantly simplifying the proof of divergence. The lower bound we obtained, $x_k > \sqrt{4\sqrt{k} - 2}$, is sufficient to show that $x_n$ grows without bound.

In this article, we have thoroughly explored the problem of proving the divergence of a recursively defined sequence. We presented a detailed solution that involves squaring the recursive relation, summing an inequality, estimating a sum using an integral, and establishing a lower bound that grows to infinity. This approach provides a robust and rigorous proof of divergence.

We also discussed alternative approaches, such as using the Stolz-Cesàro theorem, asymptotic analysis, and the Integral Comparison Test. While some of these methods did not directly lead to a proof of divergence, they offered valuable insights into the behavior of the sequence and highlighted the importance of different analytical tools in problem-solving.

Key Takeaways

1. Understanding Recursive Definitions: Recursive sequences are defined by a rule that relates a term to its preceding terms. Analyzing these sequences often involves iterative techniques, inequalities, and summation.
2. Proving Divergence: To prove that a sequence diverges, it is necessary to show that it does not converge to a finite limit. This can be achieved by demonstrating that the terms grow without bound or oscillate indefinitely.
3. Establishing Lower Bounds: Finding a lower bound for the terms of a sequence is a common strategy for proving divergence. If the lower bound grows to infinity, then the sequence must also diverge.
4. Summing Inequalities: Summing inequalities over a range of indices can reveal important relationships and help establish lower or upper bounds for the sequence.
5. Integral Comparison: Comparing a sum to an integral is a powerful technique for estimating the sum's behavior. The Integral Comparison Test can be used to determine the convergence or divergence of series.
6. Stolz-Cesàro Theorem: The Stolz-Cesàro theorem is a valuable tool for evaluating limits of sequences and can be particularly useful when dealing with recursive definitions. However, it is not always the most direct method for proving divergence.
7. Asymptotic Analysis: Understanding the asymptotic behavior of a sequence can provide intuition about its growth rate and guide the development of a formal proof. However, asymptotic analysis alone is not a rigorous proof.
8. Problem-Solving Strategies: Approaching a mathematical problem often involves making initial observations, formulating a strategy, applying relevant techniques, and carefully analyzing the results. Flexibility and creativity are essential in problem-solving.

This article has provided a comprehensive exploration of a challenging mathematical problem and offered valuable insights into the techniques and principles used in proving divergence. By mastering these concepts and strategies, students, educators, and enthusiasts can enhance their problem-solving skills and deepen their understanding of mathematical analysis. The journey through this problem serves as a testament to the beauty and power of mathematical reasoning.