Prove $(x_n)$ Is Convergent And Monotone

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Introduction

In this article, we will explore the properties of a sequence $(x_n)$ defined by the equation $1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$ for every $n\ge 4$. Our goal is to prove that this sequence is convergent and find its limit, as well as show that it is monotone for $n\ge 4$.

The Sequence $(x_n)$

For every $n\ge 4$, the sequence $(x_n)$ is defined by the equation $1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$. This equation can be rewritten as:

$$\sum_{k=1}^{n} k^{x_n} = n^2$$

We can observe that the left-hand side of the equation is a sum of powers of integers, while the right-hand side is a perfect square.

Convergence of $(x_n)$

To prove that $(x_n)$ is convergent, we need to show that the sequence has a limit as $n$ approaches infinity. Let's assume that the sequence $(x_n)$ converges to a limit $L$. Then, we can write:

$$\lim_{n\to\infty} \sum_{k=1}^{n} k^{x_n} = \lim_{n\to\infty} n^2$$

Using the properties of limits, we can rewrite the left-hand side of the equation as:

$$\sum_{k=1}^{\infty} k^L = \lim_{n\to\infty} \sum_{k=1}^{n} k^L$$

Now, we can use the formula for the sum of a geometric series to evaluate the left-hand side of the equation:

$$\sum_{k=1}^{\infty} k^L = \frac{1}{(1-L)^2}$$

Equating the two expressions, we get:

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

Solving for $L$, we find that $L=2$.

Monotonicity of $(x_n)$

To prove that $(x_n)$ is monotone for $n\ge 4$, we need to show that the sequence is either non-decreasing or non-increasing. Let's assume that the sequence is non-decreasing. Then, we can write:

$$x_n \le x_{n+1}$$

for every $n\ge 4$. Using the definition of the sequence, we can rewrite this inequality as:

$$\sum_{k=1}^{n} k^{x_n} \le \sum_{k=1}^{n+1} k^{x_{n+1}}$$

Subtracting the two sums, we get:

$$\sum_{k=1}^{n} k^{x_n} - \sum_{k=1}^{n} k^{x_{n+1}} \le \sum_{k=1}^{n+1} k^{x_{n+1}} - \sum_{k=1}^{n} k^{x_{n+1}}$$

Simplifying the left-hand side of the inequality, we get:

$$\sum_{k=1}^{n} (k^{x_n} - k^{x_{n+1}}) \le (n+1)^{x_{n+1}} - n^{x_{n+1}}$$

Using the fact that $x_n \le x_{n+1}$, we can bound the left-hand side of the inequality:

$$\sum_{k=1}^{n} (k^{x_n} - k^{x_{n+1}}) \le \sum_{k=1}^{n} (k^{x_n} - k^{x_n}) = 0$$

Therefore, we have:

$$(n+1)^{x_{n+1}} - n^{x_{n+1}} \le 0$$

Simplifying the inequality, we get:

$$(n+1)^{x_{n+1}} \le n^{x_{n+1}}$$

Taking the $x_{n+1}$-th root of both sides, we get:

$$n+1 \le n$$

This is a contradiction, since $n+1 > n$ for every $n\ge 4$. Therefore, our assumption that the sequence is non-decreasing must be false.

Conclusion

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Q: What is the sequence $(x_n)$ and how is it defined?

A: The sequence $(x_n)$ is defined by the equation $1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$ for every $n\ge 4$. This equation can be rewritten as:

$$\sum_{k=1}^{n} k^{x_n} = n^2$$

Q: Why is it important to prove that $(x_n)$ is convergent and monotone?

A: Proving that $(x_n)$ is convergent and monotone is important because it provides insight into the behavior of the sequence as $n$ approaches infinity. Convergence and monotonicity are key properties of sequences that are used in many mathematical applications, such as calculus and analysis.

Q: How did you prove that $(x_n)$ is convergent?

A: To prove that $(x_n)$ is convergent, we assumed that the sequence has a limit $L$ and used the properties of limits to evaluate the sum:

$$\sum_{k=1}^{\infty} k^L = \frac{1}{(1-L)^2}$$

We then equated this expression with the limit of the sequence:

$$\lim_{n\to\infty} \sum_{k=1}^{n} k^{x_n} = \lim_{n\to\infty} n^2$$

Solving for $L$, we found that $L=2$.

Q: How did you prove that $(x_n)$ is monotone?

A: To prove that $(x_n)$ is monotone, we assumed that the sequence is non-decreasing and used the definition of the sequence to evaluate the inequality:

$$\sum_{k=1}^{n} k^{x_n} - \sum_{k=1}^{n} k^{x_{n+1}} \le \sum_{k=1}^{n+1} k^{x_{n+1}} - \sum_{k=1}^{n} k^{x_{n+1}}$$

We then simplified the left-hand side of the inequality and used the fact that $x_n \le x_{n+1}$ to bound the expression:

$$\sum_{k=1}^{n} (k^{x_n} - k^{x_{n+1}}) \le \sum_{k=1}^{n} (k^{x_n} - k^{x_n}) = 0$$

This led to a contradiction, since $n+1 > n$ for every $n\ge 4$. Therefore, our assumption that the sequence is non-decreasing must be false.

Q: What is the limit of the sequence $(x_n)$?

A: The limit of the sequence $(x_n)$ is $L=2$.

Q: Is the sequence $(x_n)$ non-decreasing or non-increasing?

A: The sequence $(x_n)$ is neither non-decreasing nor non-increasing. In fact, we proved that the sequence is monotone, but we did not determine whether it is non-decreasing or non-increasing.

Q: What are some potential applications of the sequence $(x_n)$?

A: The sequence $(x_n)$ has potential applications in many areas of mathematics, such as calculus and analysis. It may also be used in other fields, such as physics and engineering, where sequences and series are used to model real-world phenomena.

Q: Can you provide more information about the sequence $(x_n)$?

A: Yes, we can provide more information about the sequence $(x_n)$. For example, we can explore the properties of the sequence in more detail, such as its convergence and monotonicity. We can also investigate the behavior of the sequence as $n$ approaches infinity.