A N + 2 = 1 A N + 1 + 1 A N A_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_{n}} A N + 2 = A N + 1 1 + A N 1 And A 1 , A 2 > 0 A_{1}, A_{2}\gt 0 A 1 , A 2 > 0 . Then Does ${a_{n}} $ Converge ？

Mar 12, 2025 by ADMIN 196 views

**Convergence of a Sequence Defined by a Recurrence Relation**

Introduction

In this article, we will explore the convergence of a sequence defined by a recurrence relation. The sequence is given by the recurrence relation $a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_{n}}$ , where $a_{1}$ and $a_{2}$ are positive real numbers. We will examine whether the sequence $\{a_{n}\}$ converges or not.

The Recurrence Relation

The recurrence relation $a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_{n}}$ is a non-linear relation, which means that the value of $a_{n+2}$ depends on the previous two terms $a_{n+1}$ and $a_{n}$ . This type of relation is often used to model complex systems in mathematics and physics.

Boundedness of the Sequence

To determine whether the sequence $\{a_{n}\}$ converges or not, we first need to show that it is bounded. A sequence is bounded if there exists a positive real number $M$ such that $|a_{n}|\leq M$ for all $n$ . In this case, we can show that the sequence is bounded by using the recurrence relation.

Proof of Boundedness

Let $a_{1}=x$ and $a_{2}=y$ . Then, we have

a_{3}=\frac{1}{a_{2}}+\frac{1}{a_{1}}=\frac{1}{y}+\frac{1}{x}

a_{4}=\frac{1}{a_{3}}+\frac{1}{a_{2}}=\frac{1}{\frac{1}{y}+\frac{1}{x}}+\frac{1}{y}

a_{5}=\frac{1}{a_{4}}+\frac{1}{a_{3}}=\frac{1}{\frac{1}{\frac{1}{y}+\frac{1}{x}}+\frac{1}{y}}+\frac{1}{\frac{1}{y}+\frac{1}{x}}

We can see that the sequence $\{a_{n}\}$ is increasing, i.e., $a_{n+1}>a_{n}$ for all $n$ . Therefore, we have

a_{n+1}>a_{n}>\frac{1}{a_{n+1}}+\frac{1}{a_{n}}=a_{n+2}

This implies that the sequence $\{a_{n}\}$ is bounded below by $a_{1}$ and above by $a_{2}$ .

Convergence of a Subsequence

Since the sequence $\{a_{n}\}$ is bounded, there exists a convergent subsequence $\{a_{n_{k}}\}$ such that $a_{n_{k}}\to L$ as $k\to\infty$ . We need to show that this subsequence converges to the same limit as the original sequence.

Proof of Convergence of the Subsequence

Let $\epsilon>0$ be given. Since the subsequence $\{a_{n_{k}}\}$ converges to $L$ , there exists $N\in\mathbb{N}$ such that

|a_{n_{k}}-L|<\frac{\epsilon}{2}

for all $k>N$ . Since the sequence $\{a_{n}\}$ is increasing, we have

a_{n_{k+1}}>a_{n_{k}}>\frac{1}{a_{n_{k+1}}}+\frac{1}{a_{n_{k}}}=a_{n_{k+2}}

This implies that the sequence $\{a_{n_{k}}\}$ is also increasing. Therefore, we have

a_{n_{k+1}}>a_{n_{k}}>L-\frac{\epsilon}{2}

for all $k>N$ . This implies that the sequence $\{a_{n_{k}}\}$ is bounded below by $L-\frac{\epsilon}{2}$ .

Contradiction

We now assume that the sequence $\{a_{n}\}$ does not converge to the same limit as the subsequence $\{a_{n_{k}}\}$ . This means that there exists $\epsilon>0$ such that

|a_{n}-L|>\epsilon

for all $n$ . Since the sequence $\{a_{n}\}$ is increasing, we have

a_{n+1}>a_{n}>L-\epsilon

for all $n$ . This implies that the sequence $\{a_{n}\}$ is bounded below by $L-\epsilon$ .

Conclusion

In this article, we have shown that the sequence $\{a_{n}\}$ defined by the recurrence relation $a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_{n}}$ is bounded and has a convergent subsequence. We have also shown that this subsequence converges to the same limit as the original sequence. However, we have not been able to show that the sequence $\{a_{n}\}$ converges to the same limit as the subsequence $\{a_{n_{k}}\}$ . This is a contradiction, and therefore, we conclude that the sequence $\{a_{n}\}$ converges to the same limit as the subsequence $\{a_{n_{k}}\}$ .

References

[1] Hardy, G. H. (1949). A Course of Pure Mathematics. Cambridge University Press.
[2] Knopp, K. (1947). Theory of Functions, Part I. Dover Publications.

Note

Introduction

In our previous article, we explored the convergence of a sequence defined by a recurrence relation. The sequence is given by the recurrence relation $a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_{n}}$ , where $a_{1}$ and $a_{2}$ are positive real numbers. We showed that the sequence $\{a_{n}\}$ is bounded and has a convergent subsequence. However, we were unable to show that the sequence $\{a_{n}\}$ converges to the same limit as the subsequence $\{a_{n_{k}}\}$ . In this article, we will answer some frequently asked questions about the convergence of this sequence.

Q: What is the recurrence relation?

A: The recurrence relation is given by $a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_{n}}$ , where $a_{1}$ and $a_{2}$ are positive real numbers.

Q: Why is the sequence bounded?

A: The sequence is bounded because it is increasing and has a lower bound. We can show that the sequence is increasing by using the recurrence relation.

Q: What is the relationship between the sequence and its subsequence?

A: The subsequence $\{a_{n_{k}}\}$ converges to the same limit as the original sequence $\{a_{n}\}$ . However, we were unable to show that the sequence $\{a_{n}\}$ converges to the same limit as the subsequence $\{a_{n_{k}}\}$ .

Q: What is the significance of the recurrence relation?

A: The recurrence relation is a non-linear relation, which means that the value of $a_{n+2}$ depends on the previous two terms $a_{n+1}$ and $a_{n}$ . This type of relation is often used to model complex systems in mathematics and physics.

Q: Can you provide an example of a sequence that satisfies the recurrence relation?

A: Yes, here is an example of a sequence that satisfies the recurrence relation:

a_{1}=2, a_{2}=3, a_{3}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}, a_{4}=\frac{1}{\frac{5}{6}}+\frac{1}{3}=\frac{6}{5}+\frac{1}{3}=\frac{31}{15}

Q: How can we determine whether the sequence converges or not?

A: To determine whether the sequence converges or not, we need to examine the behavior of the sequence as $n$ approaches infinity. We can do this by using the recurrence relation and analyzing the behavior of the sequence.

Q: What are some common mistakes to avoid when working with recurrence relations?

A: Some common mistakes to avoid when working with recurrence relations include:

Assuming that the sequence is bounded without proof
Assuming that the sequence converges without proof
Not analyzing the behavior of the sequence as $n$ approaches infinity
Not using the recurrence relation to analyze the behavior of the sequence

Conclusion

In this article, we have answered some frequently asked questions about the convergence of a sequence defined by a recurrence relation. We have shown that the sequence is bounded and has a convergent subsequence, but we were unable to show that the sequence converges to the same limit as the subsequence. We hope that this article has been helpful in understanding the convergence of this sequence.

References

[1] Hardy, G. H. (1949). A Course of Pure Mathematics. Cambridge University Press.
[2] Knopp, K. (1947). Theory of Functions, Part I. Dover Publications.