Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $\gamma > 0$ such that $J_\alpha > 0$ for all $\alpha \in ]0,\gamma[$. Suppose that $\alpha < \gamma$. Show that $$\lim _ { n \rightarrow + \infty } \left| f _ { \alpha } ( 1 ) - R _ { n } ( 1 ) \right| = + \infty.$$

An application $h$ from a non-trivial interval $J$ of $\mathbf{R}$ to $\mathbf{R}$ is said to be log-convex if, and only if, it takes values in $\mathbf{R}_+^*$ and $\ln \circ h$ is convex on $J$.
More generally, determine, if $T \in \mathbf{R}_+^*$, all applications $g$ from $]-T, +\infty[$ to $\mathbf{R}$, log-convex and satisfying $$\forall t \in ]-T, +\infty[, (t+T)g(t) = (t+2T)g(t+2T).$$

An application $h$ from a non-trivial interval $J$ of $\mathbf{R}$ to $\mathbf{R}$ is said to be log-convex if, and only if, it takes values in $\mathbf{R}_+^*$ and $\ln \circ h$ is convex on $J$.
Does there exist an application $h$, from $\mathbf{R}$ to $\mathbf{R}$ and log-convex, satisfying $$\forall t \in \mathbf{R}, (t+T)h(t) = (t+2T)h(t+2T)?$$

Verify that we have $S(0) = \operatorname{Ent}_{\varphi}(f)$ and $\lim_{t \rightarrow +\infty} S(t) = 0$, where $S(t) = \operatorname{Ent}_{\varphi}\left(P_{t}(f)\right)$.

Verify that we have $S(0) = \operatorname{Ent}_{\varphi}(f)$ and $\lim_{t \rightarrow +\infty} S(t) = 0$.

Consider a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of real numbers strictly greater than $-1$, convergent with limit zero. Show that: $$\lim_{n \rightarrow +\infty} \sum_{i=1}^{n-1} \frac{|\varepsilon_i|}{\sqrt{i(n-i)}} = 0.$$

Consider a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of real numbers strictly greater than $-1$, convergent with limit zero. Deduce that: $$\lim_{n \rightarrow +\infty} \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}} \cdot \left(\frac{(1+\varepsilon_i)(1+\varepsilon_{n-i})}{1+\varepsilon_n} - 1\right) = 0.$$

We fix $\boldsymbol{x}, \boldsymbol{y} \in \mathscr{E}_{d}^{n}(\mathbb{R})$ and $J(\tau, R) = \sum_{i=1}^{n} |\boldsymbol{y}_{i} - (R\boldsymbol{x}_{i} + \tau)|^{2}$. For all $R \in \mathrm{SO}_{d}(\mathbb{R})$, $\tau(R)$ denotes the unique minimizer of $\tau \mapsto J(\tau, R)$.

• [(a)] Show that there exists $R_{*} \in \mathrm{SO}_{d}(\mathbb{R})$ such that $J(\tau(R_{*}), R_{*}) \leqslant J(\tau, R)$ for all $(\tau, R) \in \operatorname{Dep}(\mathbb{R}^{d})$.
• [(b)] Show that $R_{*}$ is not necessarily unique.

Let $f \in \mathcal { C } ^ { 0 } ( [ 0 , + \infty [ )$ and $\ell \in \mathbb { R }$. Prove that $$\left( \lim _ { x \rightarrow + \infty } f ( x + 1 ) - f ( x ) = \ell \right) \Rightarrow \left( \lim _ { x \rightarrow + \infty } \frac { f ( x ) } { x } = \ell \right)$$

The value of $\lim _ { n \rightarrow \infty } \frac { 1 ^ { 3 } + 2 ^ { 3 } + \ldots + n ^ { 3 } } { n ^ { 4 } }$ is:
(A) $\frac{3}{4}$
(B) $\frac{1}{4}$
(C) 1
(D) 4

The value of $\lim _ { n \rightarrow \infty } \frac { 1 ^ { 3 } + 2 ^ { 3 } + \ldots + n ^ { 3 } } { n ^ { 4 } }$ is:
(a) $\frac { 3 } { 4 }$
(b) $\frac { 1 } { 4 }$
(c) 1
(d) 4.

The value of $\lim _ { n \rightarrow \infty } \frac { 1 ^ { 3 } + 2 ^ { 3 } + \ldots + n ^ { 3 } } { n ^ { 4 } }$ is:
(a) $\frac { 3 } { 4 }$
(b) $\frac { 1 } { 4 }$
(c) 1
(d) 4.

The value of $\lim_{n \rightarrow \infty} \frac{1^3 + 2^3 + \ldots + n^3}{n^4}$ is:
(A) $\frac{3}{4}$
(B) $\frac{1}{4}$
(C) 1
(D) 4

The value of $\lim _ { n \rightarrow \infty } \frac { 1 ^ { 3 } + 2 ^ { 3 } + \ldots + n ^ { 3 } } { n ^ { 4 } }$ is:
(A) $\frac { 3 } { 4 }$
(B) $\frac { 1 } { 4 }$
(C) 1
(D) 4

Let us denote the fractional part of a real number $x$ by $\{ x \}$ (note: $\{ x \} = x - [ x ]$ where $[ x ]$ is the integer part of $x$ ). Then, $$\lim _ { n \rightarrow \infty } \left\{ ( 3 + 2 \sqrt { 2 } ) ^ { n } \right\}$$ (A) equals 0 .
(B) equals 1 .
(C) equals $\frac { 1 } { 2 }$.
(D) does not exist.

The limit $$\lim_{n \rightarrow \infty} \frac{2\log 2 + 3\log 3 + \cdots + n\log n}{n^2 \log n}$$ is equal to
(A) 0
(B) $1/4$
(C) $1/2$
(D) 1

For $a \in \mathbb { R }$ (the set of all real numbers), $a \neq - 1$, $$\lim _ { \mathrm { n } \rightarrow \infty } \frac { \left( 1 ^ { a } + 2 ^ { a } + \ldots + \mathrm { n } ^ { a } \right) } { ( n + 1 ) ^ { a - 1 } [ ( n a + 1 ) + ( n a + 2 ) + \ldots + ( n a + n ) ] } = \frac { 1 } { 60 }$$ Then $a =$
(A) 5
(B) 7
(C) $\frac { - 15 } { 2 }$
(D) $\frac { - 17 } { 2 }$

The sequence $\left( a _ { n } \right)$
$$a _ { n } = \begin{cases} 2 ^ { n } + 1 , & n \equiv 0 ( \bmod 2 ) \\ 2 ^ { n } - 1 , & n \equiv 1 ( \bmod 2 ) \end{cases}$$
is defined in the form. Accordingly, what is the value of the expression $\frac { a _ { 9 } - a _ { 7 } } { a _ { 8 } - 4 \cdot a _ { 6 } }$?
A) $-2 ^ { 8 }$
B) $-2 ^ { 7 }$
C) $-2 ^ { 6 }$
D) $-2 ^ { 5 }$
E) $-2 ^ { 4 }$