. Let $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ be the sequence such that $: \forall n \in \mathbb{N}^{*}, u_{n}=\frac{2 n+\cos (n)}{n \sin \left(\frac{1}{n}\right)+\sqrt{(n+1)(n+2)}}$.

Determine the limit of the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$.
. We write: $\forall n \in \mathbb{N}^{*}, u_{n}=\frac{2+\frac{\cos (n)}{n}}{\sin \left(\frac{1}{n}\right)+\sqrt{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}}$.

Since we have: $\lim _{n \rightarrow+\infty} \frac{\cos (n)}{n}=0, \lim _{n \rightarrow+\infty} \sin \left(\frac{1}{n}\right)=0$ and $\lim _{n \rightarrow+\infty} \sqrt{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)}=1$, we deduce that the sequence $\left(u_{n}\right)$ converges to 2.

Definition 6. Given a sequence $\left(u_{n}\right)$, we call the supremum and infimum of $\left(u_{n}\right)$ the quantities, if they exist: $\sup \left\{u_{n}, n \in \mathbb{N}\right\}$ and $\inf \left\{u_{n}, n \in \mathbb{N}\right\}$. The supremum, if it exists, of the set $\left\{u_{n}, n \in \mathbb{N}\right\}$ is the smallest of the upper bounds of this set, and its infimum, if it exists, is the largest of the lower bounds of this set.

Remark. Every non-empty and bounded above (resp. bounded below) subset of $\mathbb{R}$ has a finite supremum (resp. a finite infimum).

Theorem 2.

1. Every increasing and bounded above sequence converges to its supremum.
2. Every increasing and unbounded above sequence tends to $+\infty$.
3. Every decreasing and bounded below sequence converges to its infimum.
4. Every decreasing and unbounded below sequence tends to $-\infty$.

## Proof.

1. Let $\left(u_{n}\right)$ be an increasing and bounded above sequence.

The set $\left\{u_{n}, n \in \mathbb{N}\right\} \subset \mathbb{R}$ is bounded above, so it has a finite supremum, which we denote by $\ell$. Since $\ell$ is the smallest of the upper bounds of the sequence $\left(u_{n}\right)$, for all $\varepsilon>0, \ell-\varepsilon$ is not an upper bound of the sequence $\left(u_{n}\right)$, so there exists $N_{\varepsilon} \in \mathbb{N}$ such that: $u_{N_{\varepsilon}}>\ell-\varepsilon$. Furthermore, the sequence $\left(u_{n}\right)$ is increasing, so $\forall n \geq N_{\varepsilon}, u_{n} \geq u_{N_{\varepsilon}}$. We finally have, $\forall \varepsilon>0, \exists N_{\varepsilon} \in \mathbb{N}, \forall n \geq N_{\varepsilon}, \ell-\varepsilon < u_{n} \leq \ell$.

The sequence $\left(v_{n}\right)$ is increasing, so $\forall n \geq n_{0}, v_{n} \geq v_{n_{0}}$ and thus $\forall n \geq n_{0}, v_{n}>A$. This proves that the sequence $\left(v_{n}\right)$ tends to $+\infty$. Points 3 and 4 are proven by applying points 1 and 2 and are left as exercises.

Definition 7. Let $\left(u_{n}\right)$ and $\left(v_{n}\right)$ be two real sequences. We say that the sequences $\left(u_{n}\right)$ and $\left(v_{n}\right)$ are adjacent if:

1. $\left(u_{n}\right)$ is increasing.
2. $\left(v_{n}\right)$ is decreasing.
3. $\lim _{n \rightarrow+\infty}\left(v_{n}-u_{n}\right)=0$.

Property 2. Two adjacent sequences converge to the same limit.

Proof. $\left(u_{n}\right)$ is increasing and $\left(v_{n}\right)$ is decreasing, so the sequence $\left(v_{n}-u_{n}\right)$ is decreasing. Furthermore, the sequence $\left(v_{n}-u_{n}\right)$ converges to 0, so $\forall n \in \mathbb{N}$, $v_{n}-u_{n} \geq 0$. Therefore, we have: $u_{0} \leq u_{n} \leq v_{n} \leq v_{0}$.

The sequence $\left(u_{n}\right)$ is increasing and bounded above by $v_{0}$, so it converges, and the sequence $\left(v_{n}\right)$ is decreasing and bounded below by $u_{0}$, so it converges. Since $\lim _{n \rightarrow+\infty}\left(v_{n}-u_{n}\right)=0$, we have: $\lim _{n \rightarrow+\infty} u_{n}=\lim _{n \rightarrow+\infty} v_{n}$.