Limit of $x_n$ Sequence: $\pi/2$

In summary, the sequence $x_n$ defined by $x_{n+1}=(-1)^n(\frac{\pi}{2}-\arctan(\frac{1}{x_n}))$ is decreasing and bounded, therefore it converges to 0. This is because $|x_n|$ is equal to the decreasing sequence $y_n=\arctan(y_{n-1})$ with $y_0=1$. Therefore, the limit of $x_n$ is 0, which eliminates the first three options and leaves the last option as the correct answer.
  • #1
Vali
48
0
Let $x_{0}=1$ and $x_{n+1}=(-1)^{n}(\frac{\pi }{2}-\arctan(\frac{1}{x_{n}}))$

I have the following options to choose from:

1. $x_n$ is unbounded
2. $x_n$ is increasing and the limit of $x_n$ is $1$
3. the limit of $x_n$ is $\pi/2$.
4. the limit of $x_n$ is $0$

My attempt:

I used $$\arctan(x)+\arctan(\frac{1}{x})=\frac{\pi }{2}$$ so
$$
x_{n+1} = (-1)^n \arctan(x_n)
= \begin{cases}
\arctan(x_n) & \text{ if } n=2k \\
-\arctan(x_n) & \text{ if } n=2k+1
\end{cases}
$$

I think I should take $y_{0}=1$ and $y_{n+1}=\arctan(y_{n})$

$y_{0}=1;y_{1}=pi/4$ so the sequence is decreasing.How to find the last term? I mean how long it decreasing ?It should be
$$
\lim_{n\to \infty} \arctan(y_{n})=\frac{\pi }{2},
$$
right?

How to approach this exercise?There is another way to find the monotony of the sequence?How to find the lower limit of the sequence?
If I would show that $y_{n}$ from my question is decreasing and it's bounded then it's convergent.Then, if I note the limit of $y_{n}$ with $L$ I would get $L=arctan(L)$ so $L=0$ so the initial sequence has the limit $0$ too.
 
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  • #2
Vali said:
I think I should take $y_{0}=1$ and $y_{n+1}=\arctan(y_{n})$
You are right. Since $\arctan(-x)=-\arctan(x)$, it is easy to see that $|x_n|=y_n$ for all $n$.

You are also right that $y_n$ is decreasing. Indeed, $y_1=\pi/4<1=y_0$, and $\arctan$ is strictly increasing. Therefore, $y_2=\arctan(y_1)<\arctan(y_0)=y_1$, $y_3=\arctan(y_2)<\arctan(y_1)=y_2$ and so on.

So $|x_n|$ is decreasing, and this rules out the first three answers. And, finally, you are right that the $L=\lim_{n\to\infty}y_n$ must satisfy $L=\arctan(L)$, i.e., $L$ must be 0, and the same is true for $\lim_{n\to\infty}x_n$.
 

FAQ: Limit of $x_n$ Sequence: $\pi/2$

1. What is the limit of the sequence $x_n = \pi/2$?

The limit of the sequence $x_n = \pi/2$ is $\pi/2$, as $n$ approaches infinity. This means that as the values of $n$ increase, the terms of the sequence get closer and closer to $\pi/2$.

2. How do you prove that the limit of the sequence $x_n = \pi/2$ is $\pi/2$?

To prove that the limit of the sequence $x_n = \pi/2$ is $\pi/2$, we can use the definition of a limit. This means showing that for any given value of $\epsilon > 0$, there exists a corresponding value of $N$ such that for all $n > N$, $|x_n - \pi/2| < \epsilon$. We can also use the squeeze theorem, as the sequence is bounded between two convergent sequences: $x_n = \pi/2$ and $x_n = \pi/2 - 1/n$.

3. Is the limit of the sequence $x_n = \pi/2$ unique?

Yes, the limit of the sequence $x_n = \pi/2$ is unique. This means that no matter how we approach the limit (from the left or the right), the value will always be $\pi/2$. This is a property of limits and is known as the uniqueness of limits.

4. Can the limit of the sequence $x_n = \pi/2$ be any other value?

No, the limit of the sequence $x_n = \pi/2$ cannot be any other value. This is because the sequence is defined as $x_n = \pi/2$ for all $n$, so the limit will always be $\pi/2$. If the sequence were defined differently, then the limit could potentially be a different value.

5. How does the limit of the sequence $x_n = \pi/2$ relate to the value of pi?

The limit of the sequence $x_n = \pi/2$ does not necessarily have a direct relationship to the value of pi. However, we can see that as $n$ approaches infinity, the terms of the sequence get closer and closer to $\pi/2$, which is half of the value of pi. This can be seen as a way of approximating the value of pi, but it is not a direct relationship.

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