- #1

Vali

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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.