(a) Show that $$\underline{M}(n) \geqslant \frac{n^{2}}{2^{n-1}} \binom{n-1}{\left\lfloor \frac{n}{2} \right\rfloor}.$$
(b) Show next, using Stirling's formula recalled in the preamble, that this lower bound is equivalent to $C n^{\alpha}$ as $n$ tends to infinity, for constants $C$ and $\alpha > 0$ that one will make explicit. Compare with the upper bound for $\underline{M}(n)$ obtained in question 6 of Part II.

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.

Recall Stirling's formula, then determine a real number $c > 0$ such that $$\binom { 2 n } { n } \underset { n \rightarrow + \infty } { \sim } c \frac { 4 ^ { n } } { \sqrt { n } }$$

If $\alpha$ is an element of $]0,1[$, show, for example by using a series-integral comparison, that $$\sum _ { k = 1 } ^ { n } \frac { 1 } { k ^ { \alpha } } \underset { n \rightarrow + \infty } { \sim } \frac { n ^ { 1 - \alpha } } { 1 - \alpha }$$ If $\alpha$ is an element of $]1 , + \infty[$, show similarly that $$\sum _ { k = n + 1 } ^ { + \infty } \frac { 1 } { k ^ { \alpha } } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { ( \alpha - 1 ) n ^ { \alpha - 1 } }$$

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 $$a _ { n } \underset { n \rightarrow + \infty } { \sim } \frac { 1 } { C \ln ( n ) }$$

We admit that $P \left( e ^ { - t } \right) \sim \sqrt { \frac { t } { 2 \pi } } \exp \left( \frac { \pi ^ { 2 } } { 6 t } \right)$ as $t$ tends to $0 ^ { + }$.
By applying formula (1) to $t = \frac { \pi } { \sqrt { 6 n } }$, prove that
$$p _ { n } \sim \frac { \exp \left( \pi \sqrt { \frac { 2 n } { 3 } } \right) } { 4 \sqrt { 3 } n } \quad \text { as } n \rightarrow + \infty$$

Taking $t = \frac { \pi } { \sqrt { 6 n } }$ in the formula
$$p _ { n } = \frac { e ^ { n t } P \left( e ^ { - t } \right) } { 2 \pi } \int _ { - \pi } ^ { \pi } e ^ { - i n \theta } \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \mathrm { d } \theta$$
conclude that
$$p _ { n } = O \left( \frac { \exp \left( \pi \sqrt { \frac { 2 n } { 3 } } \right) } { n } \right) \quad \text { when } n \text { tends to } + \infty$$

By taking $t = \frac{\pi}{\sqrt{6n}}$ in formula $$p_n = \frac{e^{nt} P(e^{-t})}{2\pi} \int_{-\pi}^{\pi} e^{-in\theta} \frac{P(e^{-t}e^{i\theta})}{P(e^{-t})} \mathrm{d}\theta,$$ conclude that $$p_n = O\left(\frac{\exp\left(\pi\sqrt{\frac{2n}{3}}\right)}{n}\right) \quad \text{when } n \text{ tends to } +\infty.$$

For natural number $n$, we set $r_n = \sum_{k=n+1}^{+\infty} \frac{1}{k!}$. Prove the bound $$r_n \leq \frac{1}{(n+1)!} \sum_{k=0}^{+\infty} \frac{1}{(n+2)^k}$$ Deduce a simple equivalent of $r_n$ as $n$ tends to $+\infty$.

By continuing to bound the right-hand side of the equality in question 13, establish the estimate $$d_{VT}\left(p_{X_n}, \pi_1\right) \underset{n \rightarrow +\infty}{=} O\left(\frac{2^n}{(n+1)!}\right)$$ One may use binomial coefficients.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. We introduce the set $I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$ and assume that $n$ and $k$ vary such that $k \in I _ { n }$.
Show that we have $$k ! ( n - k ) ! = 2 \pi \mathrm { e } ^ { - n } k ^ { k + 1 / 2 } ( n - k ) ^ { n - k + 1 / 2 } \left( 1 + O \left( \frac { 1 } { n } \right) \right)$$ as $n$ tends to infinity. One may use Stirling's formula: $n ! = \left( \frac { n } { \mathrm { e } } \right) ^ { n } \sqrt { 2 \pi n } \left( 1 + O \left( \frac { 1 } { n } \right) \right)$.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19. We assume $k \in I _ { n } = \left\{ k \in \llbracket 0 , n \rrbracket \mid x _ { n , k } \in [ 0 , \ell + 1 ] \right\}$.
Deduce that, as $n$ tends to $+ \infty$, we have $$B _ { n } \left( x _ { n , k } \right) = \frac { 1 } { \sqrt { 2 \pi } } \frac { 1 + O \left( \frac { 1 } { n } \right) } { \left( \frac { 2 k } { n } \right) ^ { k + 1 / 2 } \left( 2 - \frac { 2 k } { n } \right) ^ { n - k + 1 / 2 } }$$

Show that when $n$ tends to $+\infty$, we have an equivalent of the form: $$\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \underset{n \to +\infty}{\sim} \lambda \sqrt{n},$$ where the constant $\lambda$ is to be determined.

Let $(c_n)_{n \in \mathbf{N}^*}$ and $(d_n)_{n \in \mathbf{N}^*}$ be two sequences of strictly positive real numbers such that: $c_n \underset{n \to +\infty}{\sim} d_n$ and the series $\sum_n c_n$ diverges.
We admit without proof the following result:
Theorem 1. Let $(a_n)_{n \in \mathbf{N}^*}$ and $(b_n)_{n \in \mathbf{N}^*}$ be two sequences of nonzero real numbers such that $a_n = o(b_n)$ as $n \to +\infty$ and the series $\sum_n |b_n|$ is divergent. Then: $$\sum_{k=1}^n a_k = o\!\left(\sum_{k=1}^n |b_k|\right) \text{ as } n \to +\infty.$$
By using this result, show that the series $\sum_n d_n$ is divergent and that: $$\sum_{k=1}^n c_k \underset{n \rightarrow +\infty}{\sim} \sum_{k=1}^n d_k.$$

We consider a sequence of random variables $(X_n : \Omega \longrightarrow \{-1,1\})_{n \in \mathbf{N}}$ defined on the same probability space $(\Omega, \mathscr{A}, P)$, taking values in $\{-1,1\}$, mutually independent and centered. The random variable $N_n$ counts the number of equality indices of the path $(X_1(\omega), \cdots, X_{2n}(\omega))$, and it has been shown that: $$\mathbb{E}(N_n) = \sum_{i=1}^n \frac{\binom{2i}{i}}{4^i}.$$ Deduce the equivalent: $$\mathbb{E}(N_n) \underset{n \to +\infty}{\sim} \frac{2}{\sqrt{\pi}} \sqrt{n}.$$

In an urn containing $n$ white balls and $n$ black balls, we proceed to draw balls without replacement, until the urn is completely empty. The random variable $M_n$ counts the number of equality indices $k$ between $1$ and $2n$, and it has been shown that: $$\mathbb{E}(M_n) = \sum_{i=0}^{n-1} \frac{\binom{2i}{i} \cdot \binom{2n-2i}{n-i}}{\binom{2n}{n}}.$$ Deduce the equivalent: $$\mathbb{E}(M_n) \underset{n \to +\infty}{\sim} \sqrt{\pi n}.$$

By comparison with an integral, establish that $$\sum_{k=1}^{n} \ln(k) \underset{n \rightarrow +\infty}{=} n\ln(n) - n + O(\ln(n))$$

For all real $t \geqslant 2$, we define $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Establish that $\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} \underset{n \rightarrow +\infty}{=} \ln_{2}(n) + c_{1} + O\left(\frac{1}{\ln(n)}\right)$, for a real $c_{1} \in \mathbb{R}$ to be determined.

By comparison with an integral, establish that $$\sum_{k=1}^{n} \ln(k) \underset{n \rightarrow +\infty}{=} n\ln(n) - n + O(\ln(n))$$

We set, for all real $t \geqslant 2$, $$R(t) = \sum_{\substack{p \leqslant t \\ p \text{ prime}}} \frac{\ln(p)}{p} - \ln(t)$$ Establish that $\sum_{\substack{p \leqslant n \\ p \text{ prime}}} \frac{1}{p} \underset{n \rightarrow +\infty}{=} \ln_2(n) + c_1 + O\left(\frac{1}{\ln(n)}\right)$, for a real $c_1 \in \mathbb{R}$ to be determined.

By denoting $H _ { n } := \sum _ { k = 1 } ^ { n } \dfrac { 1 } { k }$ the harmonic series, show that $$H _ { n } \sim \ln n \quad ( n \rightarrow + \infty )$$

For all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, define $R _ { p , q } := \dfrac { 1 } { q } I _ { p , q }$ where $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x, \quad \alpha_{p,q} = \frac{p}{q}.$$
Using the change of variables $s = x ^ { n + 1 }$ in $I _ { p , q } ( n )$, prove that $$R _ { p , q } ( n ) \sim \frac { 1 } { 2 p n } \quad ( n \rightarrow + \infty )$$

For all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, define $R _ { p , q } := \dfrac { 1 } { q } I _ { p , q }$ where $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x, \quad \alpha_{p,q} = \frac{p}{q},$$ and recall that $$\phi _ { p , q } ( n ) = \frac { 1 } { q } \left( \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { \alpha _ { p , q } } } d x + ( - 1 ) ^ { n } I _ { p , q } ( n ) \right).$$
Using the result $R _ { p , q } ( n ) \sim \dfrac { 1 } { 2 p n }$ as $n \to +\infty$, deduce the convergence rate of the alternating congruent-harmonic series $\sum u _ { k }$, that is, that of the sequence of partial sums $\left( \phi _ { p , q } ( n ) \right) _ { n \in \mathbf { N } }$.

Find, with proof, all possible values of $t$ such that
$$\lim _ { n \rightarrow \infty } \left\{ \frac { 1 + 2 ^ { 1/3 } + 3 ^ { 1/3 } + \cdots + n ^ { 1/3 } } { n ^ { t } } \right\} = c$$
for some real number $c > 0$. Also find the corresponding values of $c$.