Prove $(x_n)$ Is Convergent And Monotone

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Introduction

In this article, we will discuss the convergence and monotonicity of a sequence $(x_n)$ defined by the equation $1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$ for $n\ge 4$. We will prove that the sequence is convergent and find its limit, and also show that it is monotone for $n\ge 4$.

Convergence of $(x_n)$

To prove the convergence of $(x_n)$, we need to show that the sequence is bounded and monotone. We will first show that the sequence is bounded.

Boundedness of $(x_n)$

Let $n\ge 4$. We can rewrite the equation as:

$$1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$$

Since $1^{x_n}=1$, we have:

$$1+2^{x_n}+...+n^{x_n}=n^2$$

Now, we can see that:

$$2^{x_n}+...+n^{x_n}\le n^2-1$$

Since $2^{x_n}\ge 2$ for all $x_n$, we have:

$$2^{x_n}+...+n^{x_n}\ge 2+...+n^{x_n}$$

Using the AM-GM inequality, we have:

$$2+...+n^{x_n}\ge n^{x_n}$$

Therefore, we have:

$$2^{x_n}+...+n^{x_n}\ge n^{x_n}$$

Combining the two inequalities, we get:

$$n^{x_n}\le 2^{x_n}+...+n^{x_n}\le n^2-1$$

This shows that the sequence $(x_n)$ is bounded below by $x_n\ge \log_2(n^2-1)$.

Monotonicity of $(x_n)$

To show that the sequence $(x_n)$ is monotone, we need to show that $x_n\le x_{n+1}$ for all $n\ge 4$.

Let $n\ge 4$. We can rewrite the equation as:

$$1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$$

$$1^{x_{n+1}}+2^{x_{n+1}}+...+(n+1)^{x_{n+1}}=(n+1)^2$$

Subtracting the two equations, we get:

$$(n+1)^{x_{n+1}}-n^{x_n}=n^2-(n+1)^2$$

Simplifying, we get:

$$(n+1)^{x_{n+1}}-n^{x_n}=n^2-2n-1$$

Now, we can see that:

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

Using the AM-GM inequality, we have:

$$(n+1)^{x_n}-n^{x_n}\ge (n+1)^{x_n}-(n+1)^{x_n-1}$$

Simplifying, we get:

$$(n+1)^{x_n}-n^{x_n}\ge (n+1)^{x_n-1}$$

Therefore, we have:

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

Since $x_n\ge \log_2(n^2-1)$, we have:

$$(n+1)^{x_n-1}\ge (n+1)^{\log_2(n^2-1)-1}$$

Simplifying, we get:

$$(n+1)^{x_n-1}\ge \frac{n^2-1}{n+1}$$

Therefore, we have:

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

Since $n^2-1\ge 2n+1$ for all $n\ge 4$, we have:

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

Simplifying, we get:

$$(n+1)^{x_{n+1}}-n^{x_n}\ge 2$$

Therefore, we have:

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

This shows that the sequence $(x_n)$ is monotone increasing.

Convergence of $(x_n)$

Since the sequence $(x_n)$ is bounded and monotone increasing, it is convergent.

Let $x=\lim_{n\to\infty}x_n$. Then, we have:

$$1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$$

Taking the limit as $n\to\infty$, we get:

$$1^x+2^x+...=\infty$$

This is a contradiction, since the sum of the terms is infinite.

Therefore, we must have:

$$x=0$$

This shows that the sequence $(x_n)$ converges to $x=0$.

Monotonicity of $(x_n)$

To show that the sequence $(x_n)$ is monotone for $n\ge 4$, we need to show that $x_n\le x_{n+1}$ for all $n\ge 4$.

Let $n\ge 4$. We can rewrite the equation as:

$$1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$$

$$1^{x_{n+1}}+2^{x_{n+1}}+...+(n+1)^{x_{n+1}}=(n+1)^2$$

Subtracting the two equations, we get:

$$(n+1)^{x_{n+1}}-n^{x_n}=n^2-(n+1)^2$$

Simplifying, we get:

$$(n+1)^{x_{n+1}}-n^{x_n}=n^2-2n-1$$

Now, we can see that:

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

Using the AM-GM inequality, we have:

$$(n+1)^{x_n}-n^{x_n}\ge (n+1)^{x_n}-(n+1)^{x_n-1}$$

Simplifying, we get:

$$(n+1)^{x_n}-n^{x_n}\ge (n+1)^{x_n-1}$$

Therefore, we have:

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

Since $x_n\ge \log_2(n^2-1)$, we have:

$$(n+1)^{x_n-1}\ge (n+1)^{\log_2(n^2-1)-1}$$

Simplifying, we get:

$$(n+1)^{x_n-1}\ge \frac{n^2-1}{n+1}$$

Therefore, we have:

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

Since $n^2-1\ge 2n+1$ for all $n\ge 4$, we have:

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

Simplifying, we get:

$$(n+1)^{x_{n+1}}-n^{x_n}\ge 2$$

Therefore, we have:

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

This shows that the sequence $(x_n)$ is monotone increasing.

Conclusion

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Introduction

In our previous article, we proved that the sequence $(x_n)$ defined by the equation $1^{x_n}+2^{x_n}+...+n^{x_n}=n^2$ for $n\ge 4$ is convergent and monotone. We also found the limit of the sequence to be $x=0$. In this article, we will answer some frequently asked questions related to the sequence $(x_n)$.

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

A: The sequence $(x_n)$ is significant because it provides a counterexample to the statement that a sequence that is bounded and monotone is necessarily convergent. This sequence shows that a sequence can be bounded and monotone, but still diverge.

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

A: We proved that the sequence $(x_n)$ is convergent by showing that it is bounded and monotone. We first showed that the sequence is bounded below by $x_n\ge \log_2(n^2-1)$. Then, we showed that the sequence is monotone increasing by using the AM-GM inequality. Finally, we showed that the sequence converges to $x=0$ by taking the limit as $n\to\infty$.

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

A: The limit of the sequence $(x_n)$ is $x=0$. This means that as $n$ approaches infinity, the value of $x_n$ approaches 0.

Q: Is the sequence $(x_n)$ monotone for all $n\ge 4$?

A: Yes, the sequence $(x_n)$ is monotone for all $n\ge 4$. We proved this by showing that $x_n\le x_{n+1}$ for all $n\ge 4$.

Q: Can you provide a visual representation of the sequence $(x_n)$?

A: Unfortunately, it is not possible to provide a visual representation of the sequence $(x_n)$ because it is a sequence of real numbers. However, we can provide a graph of the function $f(x)=1^x+2^x+...+n^x$ to illustrate the behavior of the sequence.

Q: How does the sequence $(x_n)$ relate to other mathematical concepts?

A: The sequence $(x_n)$ is related to other mathematical concepts such as the concept of convergence and the concept of monotonicity. It also relates to the concept of limits and the concept of infinite series.

Q: Can you provide more examples of sequences that are convergent and monotone?

A: Yes, there are many examples of sequences that are convergent and monotone. Some examples include the sequence $(1/n)$, the sequence $(1/n^2)$, and the sequence $(\log n)$.

Conclusion

In this article, we have answered some frequently asked questions related to the sequence $(x_n)$. We have also provided a brief overview of the significance of the sequence and its relationship to other mathematical concepts. If you have any further questions, please don't hesitate to ask.

Additional Resources

For more information on the sequence $(x_n)$ and its properties, please refer to the following resources:

• [1] "Convergence and Monotonicity of Sequences" by John Doe
• [2] "Limits and Infinite Series" by Jane Smith
• [3] "Mathematical Analysis" by Michael Brown

Note: The above resources are fictional and for demonstration purposes only.