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

[Vali]

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.

 

[Evgeny.Makarov]

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 $\lim_{n\to\infty}x_n$.