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grandes-ecoles 2025 Q10 Convergence proof and limit determination View
10. In the general case, show that $( u _ { n } )$ converges and specify its limit.
Part 3: Linear recurrent sequences with variable coefficients
We keep the notation from the previous part and we now consider a sequence $\left( v _ { n } \right) _ { n \geqslant 0 }$ satisfying a recurrence of the form
$$v _ { n + d } = \sum _ { i = 0 } ^ { d - 1 } b _ { i } ( n ) v _ { n + i }$$
where $v _ { 0 } , \ldots , v _ { d - 1 }$ are given and for all $i \in \{ 0 , \ldots , d - 1 \}$, $\left( b _ { i } ( n ) \right) _ { n \geqslant 0 }$ is a sequence with complex values converging to $a _ { i }$. We also define for all $n \geqslant 0$, $V _ { n } = \left( v _ { n } , \ldots , v _ { n + d - 1 } \right)$. We always assume hypothesis (*) is satisfied.
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 $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.
grandes-ecoles 2025 Q11 Coefficient and growth rate estimation View
Problem 2, Part 3: Linear recurrence sequences with variable coefficients
We consider a sequence $\left( v _ { n } \right) _ { n \geqslant 0 }$ satisfying a recurrence of the form $$v _ { n + d } = \sum _ { i = 0 } ^ { d - 1 } b _ { i } ( n ) v _ { n + i }$$ where $v _ { 0 } , \ldots , v _ { d - 1 }$ are given and for all $i \in \{ 0 , \ldots , d - 1 \} , \left( b _ { i } ( n ) \right) _ { n \geqslant 0 }$ is a sequence with complex values converging to $a _ { i }$. We also define for all $n \geqslant 0 , V _ { n } = \left( v _ { n } , \ldots , v _ { n + d - 1 } \right)$. We always assume hypothesis (*) is satisfied (all complex roots of $P(X) = X^d - \sum_{i=0}^{d-1} a_i X^i$ have modulus strictly less than 1), and $A$ is the matrix from question 7.
Let $\varepsilon > 0$ be fixed. Show that there exists an integer $q \geqslant 1$ and an integer $n _ { 0 }$ such that for all $n \geqslant n _ { 0 }$, $$\left\| V _ { n + q } \right\| _ { \infty } \leqslant ( \sigma ( A ) + \varepsilon ) ^ { q } \left\| V _ { n } \right\| _ { \infty }$$ where $A$ is the matrix from question 7.
grandes-ecoles 2025 Q11 Coefficient and growth rate estimation View
11. Let $\varepsilon > 0$ be fixed. Show that there exists an integer $q \geqslant 1$ and an integer $n _ { 0 }$ such that for all $n \geqslant n _ { 0 }$,
$$\left\| V _ { n + q } \right\| _ { \infty } \leqslant ( \sigma ( A ) + \varepsilon ) ^ { q } \left\| V _ { n } \right\| _ { \infty }$$
where $A$ is the matrix from question 7.
Let $k \in \mathbf { N } ^ { * }$. Using equation $(E)$ satisfied by $y$, exhibit a recurrence relation linking $b _ { k + 1 } , b _ { k }$ and $d _ { k }$.
The equation $(E)$ is $y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }$, with $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$ and $d_k$ as defined in question 11.
grandes-ecoles 2025 Q12 Convergence proof and limit determination View
Problem 2, Part 3: Linear recurrence sequences with variable coefficients
We consider a sequence $\left( v _ { n } \right) _ { n \geqslant 0 }$ satisfying a recurrence of the form $$v _ { n + d } = \sum _ { i = 0 } ^ { d - 1 } b _ { i } ( n ) v _ { n + i }$$ where $v _ { 0 } , \ldots , v _ { d - 1 }$ are given and for all $i \in \{ 0 , \ldots , d - 1 \} , \left( b _ { i } ( n ) \right) _ { n \geqslant 0 }$ is a sequence with complex values converging to $a _ { i }$. We also define for all $n \geqslant 0 , V _ { n } = \left( v _ { n } , \ldots , v _ { n + d - 1 } \right)$. We always assume hypothesis (*) is satisfied.
Deduce that $v _ { n }$ tends to 0.
Show that
$$\left\| y _ { N } - y \right\| _ { \infty , \mathbf { R } _ { + } } \leq \frac { M } { 2 ^ { N } }$$
and deduce that $y _ { N }$ converges uniformly to $y$ on $\mathbf { R } _ { + }$. Then propose an interval $J \subset \mathbf { R } _ { + }$ where the bound on $\left\| y _ { N } - y \right\| _ { \infty , J }$ would be sharper.
Here $y _ { N } ( x ) = \sum _ { n = 0 } ^ { N } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x }$ and $y ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } \mathrm { e } ^ { - \lambda _ { n } x }$ with $\left| a_n \right| \leq \frac{M}{2^n}$.
By denoting $H _ { n } := \sum _ { k = 1 } ^ { n } \dfrac { 1 } { k }$ the harmonic series, show that $$H _ { n } \sim \ln n \quad ( n \rightarrow + \infty )$$
We assume in this part that all roots of $p$ are stable and have multiplicity 1 and we denote by $h = Xp'$ (where $p'$ is the derivative polynomial of $p$) and $h_0$ the reciprocal polynomial of $h$. We recall that, according to question 3, there exists a real number $\lambda \in \{-1, 1\}$ such that $p = \lambda p_0$.
For every real number $r > 0$, we denote by $F(r) = J(p(rX))$.
Show that $$\lim_{r \rightarrow 1^-} \pi\left(\frac{n}{2(r-1)} F(r)\right) = n - \sigma(p)$$
In this subsection, we assume that $J_n = J_n^{(1)}$, the matrix introduced in subsection A-IV.
We set $A = \begin{pmatrix} \mathrm{e}^{\beta - h} & \mathrm{e}^{-\beta - h} \\ \mathrm{e}^{-\beta + h} & \mathrm{e}^{\beta + h} \end{pmatrix}$.
Show then that $$\psi_n(h) \underset{n \rightarrow +\infty}{\longrightarrow} \ln\left(\mathrm{e}^{\beta} \operatorname{ch}(h) + \sqrt{\mathrm{e}^{2\beta} \operatorname{ch}^2(h) - 2\operatorname{sh}(2\beta)}\right).$$
We assume in this part that all roots of $p$ are stable and have multiplicity 1 and we denote by $h = Xp'$ (where $p'$ is the derivative polynomial of $p$) and $h_0$ the reciprocal polynomial of $h$. We recall that, according to question 3, there exists a real number $\lambda \in \{-1, 1\}$ such that $p = \lambda p_0$.
For every real number $r > 0$, we denote by $F(r) = J(p(rX))$.
Justify that the map $F : \mathbf{R}_+^* \rightarrow S_n(\mathbf{R})$ is differentiable and that $$F'(1) = 2n(p(S))^\top p(S) - 2S^\top (p'(S))^\top p(S) - 2(p(S))^\top p'(S) S.$$
In this subsection, we assume that $J_n = J_n^{(1)}$, the matrix introduced in subsection A-IV.
Deduce an expression for the function $m$ and conclude that $m^+ = 0$.
Recall that $m = \psi'$ when $\psi$ is differentiable on $\mathbb{R}_+^*$.
We consider the general case, without having information on the stability and multiplicity of the roots of $p$, and we seek to calculate $\sigma(p)$.
We construct the two polynomials $f$ and $g$ satisfying $f = p \wedge p_0$ and $p = fg$.
Show that $\sigma(g) = \pi(J(g))$.
We consider the general case, without having information on the stability and multiplicity of the roots of $p$, and we seek to calculate $\sigma(p)$.
We construct the two polynomials $f$ and $g$ satisfying $f = p \wedge p_0$ and $p = fg$.
Propose a method allowing us to construct a finite number (possibly zero) of polynomials $g_1, \ldots, g_\ell$, whose roots are stable and of multiplicity 1, such that $f = g_1 g_2 \cdots g_\ell$. Express $\sigma(p)$ using $n$, $\deg g$, $\pi(J(g))$, $\ell$, $\pi(J(g))$ as well as $\pi(J(g_1')), \ldots, \pi(J(g_\ell'))$.
iran-konkur 2014 Q105 Summation of sequence terms View
105- What is the value of $\dfrac{t^{11}+t^{10}+t^9+\cdots+t+1}{t^4+t^2+t^2+1}$, for $t=\dfrac{-1+\sqrt{5}}{2}$?
(1) $2$ (2) $3$ (3) $4$ (4) $5$
113. For values $n \geq n_0$, if the distance of the terms of the sequence $\left\{\dfrac{fn+1}{rn-2}\right\}$ from its limit is less than $0.02$, what is the smallest value of $n_0$?
(1) $61$ (2) $62$ (3) $63$ (4) $64$
iran-konkur 2015 Q113 Multiple-choice on sequence properties View
113- What is the limit of the sequence $a_n = \left(\dfrac{n+2}{n+1}\right)^{2n+2}$ as $n \to \infty$?
p{3cm} p{3cm} p{3cm}} (4) $3e^2$(3) $3e$(2) $e^2$(1) $3e$

iran-konkur 2016 Q113 Multiple-choice on sequence properties View
113. The sequence $\left\{\dfrac{n^2 + (-1)^n}{2n^2 + 2}\right\}$ is of what type?
(1) Divergent -- divergent (2) Non-increasing -- convergent (3) Decreasing -- convergent (4) Increasing -- divergent
iran-konkur 2016 Q115 Multiple-choice on sequence properties View
115. The largest lower bound of the sequence $\left\{\dfrac{2n+1}{3n+1}\right\}$ is which of the following?
(4) $1$ (3) $\dfrac{2}{4}$ (2) $\dfrac{5}{7}$ (1) $\dfrac{2}{3}$
iran-konkur 2017 Q113 Multiple-choice on sequence properties View
113- The sequence $\left\{\left[\dfrac{(-1)^n}{n}\right]\right\}$, $n = 1, 2, 3, \ldots$ is how?
(1) Converges to $-1$ (2) Converges to zero (3) Divergent -- bounded (4) Divergent
113- For natural numbers $n \geq n_0$, the sequence $\left\{\dfrac{2n^2+1}{n^2+2n}\right\}$ converges to its limit point, with a distance less than $0.04$. The smallest value of $n_0$ is which of the following?
(1) $96$ (2) $97$ (3) $98$ (4) $99$
iran-konkur 2018 Q114 Multiple-choice on sequence properties View
114- The sequence $\left\{\left(1+\dfrac{1}{n^2}\right)^n\right\}$ converges to which number?
(1) $\sqrt{e}$ (2) $\dfrac{1}{2}e$ (3) $1$ (4) $\dfrac{1}{e}$
iran-konkur 2020 Q115 4 marks Convergence proof and limit determination View
115. The sequence $\{x_n\}$ is defined as follows. What is the limit of $\{x_n\}$?
$$x_0 = 3 \;,\quad x_{n+1} = \frac{3x_n^2 + 64}{4x_n^2} \;,\quad (n = 1, 2, \ldots)$$
$$2\sqrt{2} \quad (1) \qquad -2\sqrt{2} \quad (2) \qquad 2\sqrt[4]{2} \quad (3) \qquad -2\sqrt[4]{2} \quad (4)$$
iran-konkur 2021 Q109 Direct term computation from recurrence View
109. Suppose the general term of the logistic recurrence sequence $a_{n+1} = \dfrac{1}{a_n} + 1$ with condition $a_1 = 1$ equals $\dfrac{k}{m}$. The ninth and eighth terms of the sequence, which one is correct?
(1) $\dfrac{k-m}{2m-k}$ (2) $\dfrac{k-2m}{k-m}$ (3) $\dfrac{k-m}{k-2m}$ (4) $\dfrac{2m-k}{k-m}$
iran-konkur 2021 Q110 Summation of sequence terms View
110. The sequence $a_n = \begin{cases} 2^k & ; \ n = 2k \\ -2k+4 & ; \ n = 2k+1 \\ \left[\dfrac{n}{k+2}\right]+a & ; \ n = 2k+2 \end{cases}$ is defined for integer values of $n$, and is assumed. If the sum of the first 10 terms of this sequence is 19, then the value of $a_2 + a_5 + a_4 + \cdots + a_{29}$ is:
(1) $-2$ (2) zero (3) $2$ (4) $1$