Consider the function $S ( a )$ defined by the limit below: $$S ( a ) : = \lim _ { n \rightarrow \infty } \frac { 1 ^ { a } + 2 ^ { a } + 3 ^ { a } + \cdots + n ^ { a } } { ( n + 1 ) ^ { a - 1 } [ ( n a + 1 ) + ( n a + 2 ) + \cdots + ( n a + n ) ] }$$ Find the sum of all values $a$ such that $S ( a ) = \frac { 1 } { 60 }$.

What is the value of $\lim _ { n \rightarrow \infty } \frac { n } { \sqrt { 4 n ^ { 2 } + 1 } + \sqrt { n ^ { 2 } + 2 } }$? [2 points]
(1) 1
(2) $\frac { 1 } { 2 }$
(3) $\frac { 1 } { 3 }$
(4) $\frac { 1 } { 4 }$
(5) $\frac { 1 } { 5 }$

As shown in the figure, a square A with side length 2 and a square B with side length 1 have sides parallel to each other, and the intersection point of the two diagonals of A coincides with the intersection point of the two diagonals of B. Let R be the region of A and its interior excluding the interior of B.
For a natural number $n \geqq 2$, small squares with side length $\frac { 1 } { n }$ are drawn in R according to the following rule. (가) One side of each small square is parallel to a side of A. (나) The interiors of the small squares do not overlap with each other.
According to such rules, let $a _ { n }$ be the maximum number of small squares with side length $\frac { 1 } { n }$ that can be drawn in R. For example, $a _ { 2 } = 12$ and $a _ { 3 } = 20$. When $\lim _ { n \rightarrow \infty } \frac { a _ { 2 n + 1 } - a _ { 2 n } } { a _ { 2 n } - a _ { 2 n - 1 } } = c$, find the value of $100 c$. [4 points]

For a natural number $m$, blocks in the shape of identical cubes are stacked with 1 block in column 1, 2 blocks in column 2, 3 blocks in column 3, $\cdots$, and $m$ blocks in column $m$. The following trial is repeated until there are no columns with an even number of blocks remaining.
For each column with an even number of blocks, remove from that column a number of blocks equal to $\frac { 1 } { 2 }$ of the number of blocks in that column.
Let $f ( m )$ be the sum of the number of blocks remaining in columns 1 through $m$ after all block removal trials are completed. For example, $f ( 2 ) = 2 , f ( 3 ) = 5 , f ( 4 ) = 6$.
$$\lim _ { n \rightarrow \infty } \frac { f \left( 2 ^ { n + 1 } \right) - f \left( 2 ^ { n } \right) } { f \left( 2 ^ { n + 2 } \right) } = \frac { q } { p }$$
Find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

For a natural number $m$, there are blocks in the shape of identical cubes stacked with 1 block in column 1, 2 blocks in column 2, 3 blocks in column 3, $\cdots$, and $m$ blocks in column $m$. The following procedure is repeated until there are no columns with an even number of blocks remaining.
For each column with an even number of blocks, remove $\frac { 1 } { 2 }$ of the blocks in that column from the column.
After completing all block removal procedures, let $f ( m )$ be the sum of the number of blocks remaining in columns 1 through $m$. For example, $f ( 2 ) = 2 , f ( 3 ) = 5 , f ( 4 ) = 6$. $$\lim _ { n \rightarrow \infty } \frac { f \left( 2 ^ { n + 1 } \right) - f \left( 2 ^ { n } \right) } { f \left( 2 ^ { n + 2 } \right) } = \frac { q } { p }$$ Find the value of $p + q$. (where $p$ and $q$ are coprime natural numbers.) [4 points]

What is the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { \sqrt { 4 n ^ { 2 } + 2 n + 1 } - 2 n }$? [2 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5

We fix $x > 0$ and for all natural integers $n$, we define $F_{n}(x)$ by: $$F_{n}(x) = \ln\left(\frac{n! \, n^{x+1}}{(x+1)(x+2) \ldots (x+n+1)}\right)$$ and $$G_{n}(x) = \ln n! + (x+1)\ln n - \left(x+n+\frac{3}{2}\right)\ln(x+n+1) + n+1 + \left(x+\frac{1}{2}\right)\ln x$$
V.C.1) Using Stirling's formula, show that: $$\lim_{n \rightarrow +\infty} G_{n}(x) = \left(x+\frac{1}{2}\right)\ln x - x + \ln\sqrt{2\pi}$$
V.C.2) Deduce that: $$\ln \Gamma(x+1) = \left(x+\frac{1}{2}\right)\ln x - x + \ln\sqrt{2\pi} - \int_{0}^{+\infty} \frac{h(u)}{u+x} du$$

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{t}{e^t - 1} - 1 + \dfrac{t}{2}$ for all $t \in \mathbb{R}^{+*}$ (extended by continuity at 0).
Does $Lf(x) - \dfrac{1}{2x^2} + \dfrac{1}{x}$ admit a finite limit at $0^+$?

We approximate $\varphi _ { n } ( x )$ using partial sums
$$S _ { m } = \sum _ { p = 0 } ^ { m } ( - 1 ) ^ { p } a _ { p } \quad \text { with } \quad m \in \mathbb { N } \quad \text { and } \quad a _ { p } = \frac { 1 } { p ! ( n + p ) ! } \left( \frac { x } { 2 } \right) ^ { n + 2 p }$$
From which value $p _ { 0 }$ of $p$ is the sequence $\left( a _ { p } \right) _ { p \in \mathbb { N } }$ decreasing?

Let $p : [0,1] \rightarrow \mathbf{R}$, $x \mapsto \sqrt{|1 - 4x^{2}|}$. Determine the pointwise Hölder exponent of $p$ at $\frac{1}{2}$.
Recall: For all $f \in \mathcal{C}$ and all $x_{0} \in [0,1]$, $$\alpha_{f}(x_{0}) = \sup \{s \in [0,1[ \mid f \in \Gamma^{s}(x_{0})\} .$$

Throughout the third part, $f \in \mathcal{C}_{0}$ satisfies property $(\mathcal{P}_{1})$: there exist $x_{0} \in [0,1]$, $s \in ]0,1[$ and $c_{1} \in ]0, +\infty[$, such that for all $(j, k) \in \mathcal{I}$, $$|c_{j,k}(f)| \leq c_{1} (2^{-j} + |k 2^{-j} - x_{0}|)^{s}$$ We fix $x_{0}$, $s$, $c_{1}$ and $x \in [0,1] \backslash \{x_{0}\}$.
Show that there exists a unique $n_{0} \in \mathbf{N}$ such that $2^{-n_{0}-1} < |x - x_{0}| \leq 2^{-n_{0}}$.

Throughout the third part, $f \in \mathcal{C}_{0}$ satisfies property $(\mathcal{P}_{1})$: there exist $x_{0} \in [0,1]$, $s \in ]0,1[$ and $c_{1} \in ]0, +\infty[$, such that for all $(j, k) \in \mathcal{I}$, $$|c_{j,k}(f)| \leq c_{1} (2^{-j} + |k 2^{-j} - x_{0}|)^{s}$$ We fix $x_{0}$, $s$, $c_{1}$ and $x \in [0,1] \backslash \{x_{0}\}$. Let $n_{0}$ be the unique natural integer such that $2^{-n_{0}-1} < |x - x_{0}| \leq 2^{-n_{0}}$. We suppose that $\|f\|_{\infty} = 1$.
Show that there exists a unique $n_{1} \in \mathbf{N}$ such that $\omega_{f}(2^{-n_{1}-1}) < 2^{-n_{0} s} \leq \omega_{f}(2^{-n_{1}})$.

We assume that for all $d \geqslant 0$, $\mathbb{P}(X \in d\mathbb{Z}) < 1$, and that only finitely many $p_i$ are strictly positive. Conclude that for all $g$ of $\mathscr{F}$ piecewise continuous, $$\sum_{k=0}^{+\infty} \mathbb{E}\left(g\left(x - S_k\right)\right) \rightarrow \frac{1}{\mathbb{E}(X)} \int_{-\infty}^{+\infty} g(t)\,dt \quad \text{when} \quad x \rightarrow +\infty$$

Let $\left(u_{n}\right)_{n \in \mathbb{N}}$ be an element of $E^{c}$. Show that $\ell^{c}$ belongs to the segment $[0,1]$.

Let $k$ be a strictly positive integer and $q$ a real belonging to the interval $]0,1[$. Show that the sequences $\left(\frac{1}{(n+1)^{k}}\right)_{n \in \mathbb{N}},\left(n^{k} q^{n}\right)_{n \in \mathbb{N}}$ and $\left(\frac{1}{n !}\right)_{n \in \mathbb{N}}$ belong to $E^{c}$ and give their convergence rate.

Justify that $\zeta(x) = \sum_{n=1}^{+\infty} \frac{1}{n^x}$ admits a limit at $+\infty$.

Determine the limit of $\zeta(x)$ as $x$ tends to 1 from above.

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.

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 $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{\infty} \left(1 + \frac{k}{n}\right)^{-k} \left(\frac{1}{2}\right)^k = 1.$$

Let $\left( c _ { k } \right) _ { k \in \mathbb{N} }$ be a sequence of elements of $\mathbb{R}^{+}$ such that the power series $\sum c _ { k } x ^ { k }$ has radius of convergence 1 and the series $\sum c _ { k }$ diverges. Show that $$\sum _ { k = 0 } ^ { + \infty } c _ { k } x ^ { k } \underset { x \rightarrow 1 ^ { - } } { \longrightarrow } + \infty$$ With the element $A$ of $\mathbb{R}^{+*}$ fixed, one will show that there exists $\alpha \in ]0,1[$ such that $$\forall x \in ]1 - \alpha , 1[ , \quad \sum _ { k = 0 } ^ { + \infty } c _ { k } x ^ { k } > A$$

We call a cycle of length $k$ with values in $\llbracket 1,n \rrbracket$, any $(k+1)$-tuple $\vec{\imath} = (i_{1}, i_{2}, \ldots, i_{k}, i_{1})$ of elements of $\llbracket 1,n \rrbracket$. We denote $|\vec{\imath}|$ the number of distinct vertices of the cycle $\vec{\imath}$.
Deduce that $$\frac{1}{n^{1+k/2}} \sum_{\substack{\vec{\imath} \in \llbracket 1,n \rrbracket^{k} \\ |\vec{\imath}| \leqslant (k+1)/2}} \left|\mathbb{E}\left(X_{i_{1}i_{2}} X_{i_{2}i_{3}} \cdots X_{i_{k-1}i_{k}} X_{i_{k}i_{1}}\right)\right| \xrightarrow{n \rightarrow +\infty} 0.$$

Let $A > 2$. Show that $$\lim_{n \rightarrow +\infty} \frac{1}{n} \mathbb{E}\left(\sum_{\substack{1 \leqslant i \leqslant n \\ |\Lambda_{i,n}| \geqslant A}} |\Lambda_{i,n}|^{p}\right) = 0.$$

Deduce that $$\frac{1}{n^{1+k/2}} \sum_{\substack{\vec{\imath} \in \llbracket 1,n \rrbracket^{k} \\ |\vec{\imath}| \leqslant (k+1)/2}} \left|\mathbb{E}\left(X_{i_{1}i_{2}} X_{i_{2}i_{3}} \cdots X_{i_{k-1}i_{k}} X_{i_{k}i_{1}}\right)\right| \xrightarrow{n \rightarrow +\infty} 0.$$

Let $A > 2$. Show that $$\lim_{n \rightarrow +\infty} \frac{1}{n} \mathbb{E}\left(\sum_{\substack{1 \leqslant i \leqslant n \\ |\Lambda_{i,n}| \geqslant A}} |\Lambda_{i,n}|^{p}\right) = 0.$$

Let $s > 1$ be a real number and let $X$ be a random variable taking values in $\mathbb{N}^*$ following the zeta distribution with parameter $s$.
Show that the sequence of functions $\left(x \mapsto \prod_{k=1}^{n} p_k^{\nu_{p_k}(x)}\right)_{n \geqslant 1}$ from $\mathbb{N}^*$ to $\mathbb{N}^*$ converges pointwise to the identity function.