4. Let the sequence of real numbers $\left(x_{n}\right)_{n \geq 0}$ be such that $x_{0}=a>0$ and $x_{n+1}=x_{n}+\sqrt{1+x_{n}^{2}}, \forall n \in \mathbb{N}$.

Study the existence of the limit of the sequence $\left(y^{n} x_{n}\right)_{n \geq 1}$, where $y$ is a fixed real number.

Is it possible for the limit of the sequence $\left(y^{n} x_{n}\right)_{n \geq 1}$ to be 2015?

Dan Popescu
## Solution.

If $x_{0}=a=\operatorname{ctg} \alpha, \alpha \in\left(0, \frac{\pi}{2}\right)$, then it can be deduced inductively that $x_{n}=\operatorname{ctg} \frac{\alpha}{2^{n}}, \forall n \in \mathbb{N}$.

Then $y^{n} x_{n}=\frac{\cos \frac{\alpha}{2^{n}}}{\sin \frac{\alpha}{2^{n}}} y^{n}=\frac{\frac{\alpha}{2^{n}}}{\sin \frac{\alpha}{2^{n}}} \cdot \frac{(2 y)^{n}}{\alpha} \cdot \cos \frac{\alpha}{2^{n}}, \forall n \geq 1$.

After evaluating the standard limits, the limit $l$ of the sequence $\left(y^{n} x_{n}\right)_{n \geq 1}$ is deduced:
$l= \begin{cases}0, & \text { if } y \in\left(\frac{-1}{2}, \frac{1}{2}\right) \\ \frac{1}{\operatorname{arcctg} a}, & \text { if } y=\frac{1}{2} \\ \infty \quad, & \text { if } y \in\left(\frac{1}{2}, \infty\right)\end{cases}$

The sequence $\left(y^{n} x_{n}\right)_{n \geq 1}$ does not have a limit for $y \leq-\frac{1}{2}$.

Yes, the limit $l=2015$ is achieved for $y=\frac{1}{2}$ and $a=\operatorname{ctg} \frac{1}{2015}$.

## Grading.

| $x_{n}=\operatorname{ctg} \frac{\alpha}{2^{n}}, \forall n \in \mathbb{N}$ | $2 \mathrm{p}$ |
| :---: | :---: |
| $y^{n} x_{n}=\frac{\cos \frac{\alpha}{2^{n}}}{\sin \frac{\alpha}{2^{n}}} y^{n}=\frac{\frac{\alpha}{2^{n}}}{\sin \frac{\alpha}{2^{n}}} \cdot \frac{(2 y)^{n}}{\alpha} \cdot \cos \frac{\alpha}{2^{n}}$ | $1 \mathrm{p}$ |
| $l=\{$ |  |
| The sequence $\left(y^{n} x_{n}\right)_{n \geq 1}$ does not have a limit for $y \leq-\frac{1}{2}$ |  |
| $l=2015$ for $y=\frac{1}{2}$ and $a=\operatorname{ctg} \frac{1}{2015}$ |  |
|  |  |