We denote $\omega(t) = e^{2i\pi t}$ for $t \in [0,1]$. For all $p \in \mathbb{Z}$, we set $$I_{p} = \int_{0}^{1} \frac{\omega(t)^{p+1}}{\mathrm{e}^{\omega(t)} - 1} \,\mathrm{d}t.$$ Verify that this integral is well defined for all $p \in \mathbb{Z}$.

Using the result of Q11, deduce that for all $\zeta \in \mathbb{U}$ and all $p \in \mathbb{Z}$, $$\frac{\zeta^{p}}{\mathrm{e}^{\zeta} - 1} = \sum_{j=0}^{+\infty} (-1)^{j} \zeta^{j+p-1} \beta(\zeta)^{j}$$ where $\beta \in \mathcal{E}$ and $|\beta(\zeta)| \leqslant C < 1$ for all $\zeta \in \mathbb{U}$.

Recall that $x$ is a fixed element of $]0;1[$. Finally deduce that:
$$\forall y \in ] 0 ; \pi \left[ , \quad \sum _ { n = 1 } ^ { + \infty } \frac { 2 ( - 1 ) ^ { n } y \sin ( y ) } { y ^ { 2 } - n ^ { 2 } \pi ^ { 2 } } = 1 - \frac { \sin ( y ) } { y } . \right.$$

Show that the power series expansion of every rational function with rational coefficients is a solution of a differential equation of order 1.

We denote $\omega(t) = e^{2i\pi t}$ for $t \in [0,1]$, and for all $p \in \mathbb{Z}$, $$I_{p} = \int_{0}^{1} \frac{\omega(t)^{p+1}}{\mathrm{e}^{\omega(t)} - 1} \,\mathrm{d}t.$$ Show that $I_{0} = 1$ and that, for all $p \in \mathbb{N}^{*}$, $I_{p} = 0$.

Show that a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if there exist an integer $d \geq 0$ and polynomials not all zero $S_0, \ldots, S_d \in \mathbf{Z}[x]$ such that: for all $n \geq 0$, $$S_0(n) c_n + \cdots + S_d(n) c_{n+d} = 0.$$

Show that the power series $$h(x) = \sum_{n=0}^{\infty} \frac{(2n)!(3n)!}{(n!)^5} x^n$$ is a solution of a differential equation, then make one explicit.

For all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, the Bernoulli polynomial is defined by $$B_{n}(z) = n! \int_{0}^{1} \frac{\mathrm{e}^{z\omega(t)}}{(\mathrm{e}^{\omega(t)} - 1)\omega(t)^{n-1}} \,\mathrm{d}t$$ where $\omega(t) = e^{2i\pi t}$. Show that, for all $n \in \mathbb{N}^{*}$ and all $z \in \mathbb{C}$, $$B_{n}(z+1) - B_{n}(z) = n z^{n-1}.$$

Let $\psi$ be the function from $\mathbb{R}$ to $\mathbb{R}$ such that, for all $x \in \mathbb{R}$, $$\psi(x) = \begin{cases} \dfrac{x}{\mathrm{e}^{x} - 1} & \text{if } x \neq 0 \\ 1 & \text{otherwise} \end{cases}$$ Let furthermore $u$ be the function from $\mathbb{R}^{2}$ to $\mathbb{R}$ such that, for all $(x,t) \in \mathbb{R}^{2}$, $$u(x,t) = \psi(x)\,\mathrm{e}^{tx}.$$ Show that $u$ is of class $\mathcal{C}^{\infty}$ on $\mathbb{R}^{2}$.

Let $\psi$ be the function from $\mathbb{R}$ to $\mathbb{R}$ such that, for all $x \in \mathbb{R}$, $$\psi(x) = \begin{cases} \dfrac{x}{\mathrm{e}^{x} - 1} & \text{if } x \neq 0 \\ 1 & \text{otherwise} \end{cases}$$ Let furthermore $u$ be the function from $\mathbb{R}^{2}$ to $\mathbb{R}$ such that, for all $(x,t) \in \mathbb{R}^{2}$, $$u(x,t) = \psi(x)\,\mathrm{e}^{tx}.$$ For all $(x,t) \in \mathbb{R}^{2}$, calculate $\dfrac{\partial u}{\partial t}(x,t)$ then show that, for all $n \in \mathbb{N}^{*}$, $$\frac{\partial}{\partial t}\frac{\partial^{n} u}{\partial x^{n}}(x,t) = x\frac{\partial^{n} u}{\partial x^{n}}(x,t) + n\frac{\partial^{n-1} u}{\partial x^{n-1}}(x,t).$$

Let $\psi(x) = \begin{cases} \frac{x}{e^x-1} & x\neq 0 \\ 1 & x=0 \end{cases}$, $u(x,t) = \psi(x)e^{tx}$, and for all $n \in \mathbb{N}$, let $A_{n}$ be the function from $\mathbb{R}$ to $\mathbb{R}$ such that, for all $t \in \mathbb{R}$, $$A_{n}(t) = \frac{\partial^{n} u}{\partial x^{n}}(0,t).$$ Show that, for all $n \in \mathbb{N}$, $A_{n} = B_{n}$, where $(B_n)_{n\in\mathbb{N}}$ are the Bernoulli polynomials.

Show that $v(x)$ is not the power series expansion of a rational fraction.

Let $\varphi$ be the function defined by
$$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$
Justify that $\varphi$ is extendable by continuity at 0 and specify the value of its extension at 0. We will still denote this extension by $\varphi$.

Let $\varphi$ be the function defined by
$$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$
We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
Prove that, for all $t$ in $] - 1,1 \left[ , \varphi ^ { \prime } ( t ) = \varphi ( t ) \psi ( t ) \right.$, where
$$\forall t \in ] - 1,1 \left[ , \quad \psi ( t ) = - \sum _ { n = 0 } ^ { + \infty } \frac { 1 } { n + 2 } t ^ { n } \right.$$
then that, for all $n$ in $\mathbb { N } ^ { * }$,
$$\varphi ^ { ( n ) } ( 0 ) = - \sum _ { k = 0 } ^ { n - 1 } \frac { k ! } { k + 2 } \binom { n - 1 } { k } \varphi ^ { ( n - k - 1 ) } ( 0 )$$

Let $\varphi$ be the function defined by
$$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$
We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by
$$\left\{ \begin{array} { l } b _ { 0 } = - 1 \\ \forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k } \end{array} \right.$$
Conclude that
$$\forall t \in ] - 1,1 \left[ , \quad \varphi ( t ) = \mathrm { e } \left( 1 - \sum _ { k = 1 } ^ { + \infty } b _ { k } t ^ { k } \right) \right.$$

Show that for all $k \in \mathbf { N } ^ { * }$,
$$g ^ { ( k ) } ( 0 ) = ( - 1 ) ^ { k } d _ { k }$$
where the coefficients $d _ { k }$ are defined by
$$d _ { 0 } = 1 , \quad \text { and } \quad \forall k \geq 1 \quad d _ { k } = \sum _ { i = 1 } ^ { k } \binom { k - 1 } { i - 1 } d _ { k - i } b _ { i } ,$$
and $g \in \mathcal { C } ^ { \infty } \left( \mathbf { R } _ { + } , \mathbf { R } \right)$ is defined by $g ( x ) = \mathrm { e } ^ { y ( x ) }$, with $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ and $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.

Let $\log x = g(x) = x f(x)$. Find $f^{(n)}(1)$, the $n$-th derivative of $f$ evaluated at $x = 1$.

The limit $$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$$ (A) does not exist
(B) exists and equals 0
(C) exists and equals $2/3$
(D) exists and equals 1

The limit $$\lim _ { x \rightarrow 0 } \frac { \left( e ^ { x } - 1 \right) \tan ^ { 2 } x } { x ^ { 3 } }$$ (A) does not exist
(B) exists and equals 0
(C) exists and equals $2 / 3$
(D) exists and equals 1

If $a _ { n } = \left( 1 + \frac { 1 } { n ^ { 2 } } \right) \left( 1 + \frac { 2 ^ { 2 } } { n ^ { 2 } } \right) ^ { 2 } \left( 1 + \frac { 3 ^ { 2 } } { n ^ { 2 } } \right) ^ { 3 } \cdots \left( 1 + \frac { n ^ { 2 } } { n ^ { 2 } } \right) ^ { n }$, then $$\lim _ { n \rightarrow \infty } a _ { n } ^ { - 1 / n ^ { 2 } }$$ is
(A) 0
(B) 1
(C) $e$
(D) $\sqrt { e } / 2$

If $a _ { n } = \left( 1 + \frac { 1 } { n ^ { 2 } } \right) \left( 1 + \frac { 2 ^ { 2 } } { n ^ { 2 } } \right) ^ { 2 } \left( 1 + \frac { 3 ^ { 2 } } { n ^ { 2 } } \right) ^ { 3 } \cdots \left( 1 + \frac { n ^ { 2 } } { n ^ { 2 } } \right) ^ { n }$, then $$\lim _ { n \rightarrow \infty } a _ { n } ^ { - 1 / n ^ { 2 } }$$ is
(A) 0
(B) 1
(C) $e$
(D) $\sqrt { e } / 2$

Let $f ( x ) = \frac { 1 } { 2 } x \sin x - ( 1 - \cos x )$. The smallest positive integer $k$ such that $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x ^ { k } } \neq 0$ is:
(A) 3
(B) 4
(C) 5
(D) 6.

For all natural numbers $n$, let $$A_{n} = \sqrt{2 - \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}} \quad (n \text{ many radicals})$$
(a) Show that for $n \geq 2$, $$A_{n} = 2\sin\frac{\pi}{2^{n+1}}$$
(b) Hence, or otherwise, evaluate the limit $$\lim_{n \rightarrow \infty} 2^{n} A_{n}$$

Let $$P(x) = 1 + 2x + 7x^2 + 13x^3, \quad x \in \mathbb{R}$$ Calculate for all $x \in \mathbb{R}$, $$\lim_{n \to \infty} \left(P\left(\frac{x}{n}\right)\right)^n$$

The limit $$\lim _ { n \rightarrow \infty } n ^ { - \frac { 3 } { 2 } } \left( ( n + 1 ) ^ { ( n + 1 ) } ( n + 2 ) ^ { ( n + 2 ) } \ldots ( 2 n ) ^ { ( 2 n ) } \right) ^ { \frac { 1 } { n ^ { 2 } } }$$ equals
(A) 0.
(B) 1.
(C) $e ^ { - \frac { 1 } { 4 } }$.
(D) $4 e ^ { - \frac { 3 } { 4 } }$.