Let $f$ be the function defined by $$f(x) = \sum_{n=1}^{+\infty} \left(\frac{1}{n+x} - \frac{1}{n}\right)$$ Show that there exists $A \in \mathbb{R}_{+}^{*}$ such that $$\forall k \in \mathbb{N}^{*}, \forall x \in {]-1,1[}, \quad \left|f^{(k)}(x)\right| \leqslant k! \left(A + \frac{1}{(x+1)^{k+1}}\right)$$

Let $f$ be the function defined by $$f(x) = \sum_{n=1}^{+\infty} \left(\frac{1}{n+x} - \frac{1}{n}\right)$$ Using the result of Q18, deduce that $f$ is expandable as a power series on $]-1,1[$ and that $$\forall x \in {]-1,1[}, \quad f(x) = \sum_{k=1}^{+\infty} (-1)^{k} \zeta(k+1) x^{k}$$

Determine for which $x \in \mathbb{R}$ the integral below is convergent $$\int_{0}^{1} \frac{t^{x} - 1}{1 - t} \mathrm{~d}t$$

Let $f$ be the function defined by $$f(x) = \sum_{n=1}^{+\infty} \left(\frac{1}{n+x} - \frac{1}{n}\right)$$ By noting that, for all $t \in [0,1[$, $\frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}$, show that $$\forall x \in {]-1,+\infty[}, \quad f(x) = \int_{0}^{1} \frac{t^{x} - 1}{1 - t} \mathrm{~d}t$$

Using the results of the previous questions, deduce an integral expression of $\zeta(k+1)$ for all $k \in \mathbb{N}^{*}$.

Show that $$\forall k \in \mathbb{N}^{*}, \quad \zeta(k+1) = \frac{1}{k!} \int_{0}^{+\infty} \frac{u^{k}}{\mathrm{e}^{u} - 1} \mathrm{~d}u$$

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Show that $\zeta$ is continuous on $]1, +\infty[$.

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Bound $\sum_{n=2}^{+\infty} \frac{1}{n^s}$ by two integrals and deduce $\lim_{s \rightarrow +\infty} \zeta(s) = 1$.

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Determine $C(s)$ such that $$\forall s \in ]1, +\infty[, \quad \sum_{k=1}^{+\infty} \frac{1}{(2k-1)^s} = C(s) \zeta(s).$$

Show $$\forall n \in \mathbb{N}^{\star}, \forall s \in ]1, +\infty[, \quad \sum_{k=n+1}^{+\infty} \frac{1}{(2k-1)^s} \leqslant \frac{1}{2(s-1)} \frac{1}{(2n-1)^{s-1}}.$$

For every natural integer $n$ and every real $x$ in $J = [0, 1/2[$, set $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Justify that, for every natural integer $n$, the function $S_n$ is defined on $J$.

For every natural integer $n$ and every real $x$ in $J = [0, 1/2[$, set $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Show that the sequence $(S_n)$ converges pointwise on $J$ to the zero function.

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$, and for $x \in J = [0,1/2[$, $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Show $$\forall n \in \mathbb{N}^{\star}, \forall x \in J, \quad \pi \tan(\pi x) + S_n(x) = -\frac{2I_{4n}^{\prime}(2x)}{I_{4n}(2x)} + \frac{I_{2n}^{\prime}(x)}{I_{2n}(x)} + \sum_{p=1}^{+\infty} 2\left(2^{2p} - 1\right) \zeta(2p) x^{2p-1}.$$

Using the results of Q26 and Q28, deduce the equality $$\forall x \in J, \quad \pi \tan(\pi x) = \sum_{p=1}^{+\infty} 2\left(2^{2p} - 1\right) \zeta(2p) x^{2p-1}$$

Using the result $\alpha_{2n+1} = \frac{2(2^{2n+2}-1)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2)$ and the fact that $\lim_{s \to +\infty} \zeta(s) = 1$, deduce an equivalent of $\alpha_{2n+1}$ as $n$ tends to infinity.

Show that the series of functions $\sum u_k$ where for all $k \in \mathbb{N}^*$, the function $u_k$ is defined on $[0, +\infty[$ by $u_k : x \mapsto (1 + kx)^{-k}(1/2)^k$ is normally convergent on $[0, +\infty[$.

Deduce that for all $n \in \mathbb{N}^*, \sum_{k \geqslant 1} \left(1 + \frac{k}{n}\right)^{-k} \left(\frac{1}{2}\right)^k$ converges and that $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{\infty} \left(1 + \frac{k}{n}\right)^{-k} \left(\frac{1}{2}\right)^k = 1.$$

Show that the series $\sum P \left( S _ { n } = 0 _ { d } \right)$ is divergent if and only if $P ( R \neq + \infty ) = 1$.

For $i \in \mathbb{N}^{*}$, let $Y _ { i }$ be the Bernoulli random variable indicating the event $$Y _ { i } = \mathbf{1} \left( S _ { i } \notin \left\{ S _ { k } , 0 \leq k \leq i - 1 \right\} \right) .$$ Show that, for $i \in \mathbb{N}^{*}$: $$P \left( Y _ { i } = 1 \right) = P ( R > i )$$ Deduce that, for $n \in \mathbb{N}^{*}$: $$E \left( N _ { n } \right) = 1 + \sum _ { i = 1 } ^ { n } P ( R > i )$$

In this part, $d$ equals 1 and we simply denote $0_d = 0$. Moreover, $p$ is an element of $]0,1[$, $q = 1 - p$ and the distribution of $X$ is given by $$P ( X = 1 ) = p \quad \text{and} \quad P ( X = - 1 ) = q .$$ For $n \in \mathbb{N}$, determine $P \left( S _ { 2 n + 1 } = 0 \right)$ and justify the equality: $$P \left( S _ { 2 n } = 0 \right) = \binom { 2 n } { n } ( p q ) ^ { n }$$

In this part, $d$ equals 1 and we simply denote $0_d = 0$. Moreover, $p$ is an element of $]0,1[$, $q = 1 - p$ and the distribution of $X$ is given by $$P ( X = 1 ) = p \quad \text{and} \quad P ( X = - 1 ) = q .$$ We consider the functions $F$ and $G$ defined by the formulas $$\begin{aligned} & \forall x \in ]-1,1[ , \quad F ( x ) = \sum _ { n = 0 } ^ { + \infty } P \left( S _ { n } = 0 _ { d } \right) x ^ { n } \\ & \forall x \in [ - 1,1 ] , \quad G ( x ) = \sum _ { n = 1 } ^ { + \infty } P ( R = n ) x ^ { n } \end{aligned}$$ For $x \in ]-1,1[$, give a simple expression for $G ( x )$.
Express $P ( R = + \infty )$ as a function of $| p - q |$.
Determine the distribution of $R$.

In this part, $d$ equals 1 and we simply denote $0_d = 0$. Moreover, $p$ is an element of $]0,1[$, $q = 1 - p$ and the distribution of $X$ is given by $$P ( X = 1 ) = p \quad \text{and} \quad P ( X = - 1 ) = q .$$ We assume that $$p = q = \frac { 1 } { 2 }$$ Give a simple equivalent of $P ( R = 2 n )$ as $n$ tends to $+ \infty$. Deduce a simple equivalent of $E \left( N _ { n } \right)$ as $n$ tends to $+ \infty$.

Let $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ and $\left( b _ { n } \right) _ { n \in \mathbb{N} }$ be two sequences of elements of $\mathbb{R}^{+*}$. We assume that $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ is decreasing and that $$\forall n \in \mathbb{N} , \quad \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } = 1 .$$ We set, for $n \in \mathbb{N}$: $$B _ { n } = \sum _ { k = 0 } ^ { n } b _ { k } .$$ Let $m$ and $n$ be two natural integers such that $m > n$. Show that $$a _ { n } \leq \frac { 1 } { B _ { n } } \quad \text{and} \quad 1 \leq a _ { n } B _ { m - n } + a _ { 0 } \left( B _ { m } - B _ { m - n } \right) .$$

Let $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ and $\left( b _ { n } \right) _ { n \in \mathbb{N} }$ be two sequences of elements of $\mathbb{R}^{+*}$. We assume that $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ is decreasing and that $$\forall n \in \mathbb{N} , \quad \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } = 1 .$$ We set, for $n \in \mathbb{N}$: $$B _ { n } = \sum _ { k = 0 } ^ { n } b _ { k } .$$ We assume in this question that there exists a sequence $\left( m _ { n } \right) _ { n \in \mathbb{N} }$ satisfying $m _ { n } > n$ for $n$ large enough and $$B _ { m _ { n } - n } \underset { n \rightarrow + \infty } { \sim } B _ { n } \quad \text{and} \quad B _ { m _ { n } } - B _ { m _ { n } - n } \underset { n \rightarrow + \infty } { \longrightarrow } 0 .$$ Show that $$a _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { B _ { n } } .$$

Let $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ and $\left( b _ { n } \right) _ { n \in \mathbb{N} }$ be two sequences of elements of $\mathbb{R}^{+*}$. We assume that $\left( a _ { n } \right) _ { n \in \mathbb{N} }$ is decreasing and that $$\forall n \in \mathbb{N} , \quad \sum _ { k = 0 } ^ { n } a _ { k } b _ { n - k } = 1 .$$ We set, for $n \in \mathbb{N}$: $$B _ { n } = \sum _ { k = 0 } ^ { n } b _ { k } .$$ We assume in this question that there exists $C > 0$ such that $$b _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { C } { n }$$ Using question 17 for a well-chosen sequence $\left( m _ { n } \right) _ { n \in \mathbb{N} }$, show that $$a _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { C \ln ( n ) }$$