The question asks to prove that a sequence converges (using monotone convergence, fixed point arguments, etc.) and/or to determine its limit, typically by solving f(ℓ) = ℓ.
For a sequence $\left\{ a _ { n } \right\}$ where the sum of the first $n$ terms $S _ { n } = 2 n + \frac { 1 } { 2 ^ { n } }$, what is the value of $\lim _ { n \rightarrow \infty } a _ { n }$? [3 points] (1) 2 (2) 1 (3) $\frac { 1 } { 2 }$ (4) $\frac { 1 } { 4 }$ (5) 0
For a sequence $\left\{ a _ { n } \right\}$ with $\sum _ { n = 1 } ^ { \infty } \frac { a _ { n } } { 4 ^ { n } } = 2$, find the value of $\lim _ { n \rightarrow \infty } \frac { a _ { n } + 4 ^ { n + 1 } - 3 ^ { n - 1 } } { 4 ^ { n - 1 } + 3 ^ { n + 1 } }$. [3 points]
For a natural number $n$, let $\mathrm { P } _ { n }$ be the point where the line $x = 4 ^ { n }$ meets the curve $y = \sqrt { x }$. Let $L _ { n }$ be the length of the segment $\mathrm { P } _ { n } \mathrm { P } _ { n + 1 }$. Find the value of $\lim _ { n \rightarrow \infty } \left( \frac { L _ { n + 1 } } { L _ { n } } \right) ^ { 2 }$. [4 points]
A function $f ( x )$ satisfies $\lim _ { x \rightarrow 1 } ( x + 1 ) f ( x ) = 1$. When $\lim _ { x \rightarrow 1 } \left( 2 x ^ { 2 } + 1 \right) f ( x ) = a$, find the value of $20 a$. [3 points]
14. The three lines $l _ { 1 } : x + y - 1 = 0 , l _ { 2 } : n x + y - n = 0 , l _ { 3 } : x + n y - n = 0 \left( n \in \mathbf { N } ^ { * } , n \geq 2 \right)$ form a triangle with area denoted as $S _ { n }$ . Then $\lim _ { n \rightarrow \infty } S _ { n } =$ $\_\_\_\_$.
II. Multiple Choice (Total Score: 20 points, 5 points each)
20. (This question is worth 15 points) Given that the sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = \frac { 1 } { 2 }$ and $a _ { n + 1 } = a _ { n } - a _ { n } ^ { 2 }$ ( $n \in \mathbb{N
Let $n \in \mathbb { N }$. Show that we can order the zeros of $\varphi _ { n }$, that is, there exists a strictly increasing sequence $\left( \alpha _ { k } ^ { ( n ) } \right) _ { k \in \mathbb { N } }$ of zeros of $\varphi _ { n }$ such that $\varphi _ { n }$ does not vanish on $] 0 , \alpha _ { 0 } ^ { ( n ) } [$ and on every interval $] \alpha _ { k } ^ { ( n ) } , \alpha _ { k + 1 } ^ { ( n ) } [$ with $k$ in $\mathbb { N }$ and that $\lim _ { k \rightarrow \infty } \alpha _ { k } ^ { ( n ) } = + \infty$. Construct the sequence $\left( \alpha _ { k } ^ { ( n ) } \right) _ { k \in \mathbb { N } }$ by induction on $k$ by showing that the set $\mathcal { Z } _ { k }$ of zeros of $\varphi _ { n }$ in the interval $] \alpha _ { k } ^ { ( n ) } , + \infty [$ has a smallest element.
Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a $2 \pi$-periodic function of class $\mathcal { C } ^ { 1 }$. We consider the Fourier series of $f$ in cosines and sines, denoted $$c _ { 0 } + \sum _ { n \geq 1 } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right)$$ Deduce that, for all $t \in \mathbb { R }$, $$f ( t ) = \lim _ { x \rightarrow 1 ^ { - } } \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \frac { \left( 1 - x ^ { 2 } \right) f ( u ) } { x ^ { 2 } - 2 x \cos ( t - u ) + 1 } \mathrm { ~d} u$$
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. Show that the equation $f(x)=x$ admits a smallest solution. In what follows, we denote it by $x_f$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$ and by $l$ the limit of $(u_n)$. Show that $l=x_f$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We denote by $x_f$ the smallest solution of $f(x)=x$. We assume $m>1$. Show that $x_f\in[0,1[$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. We denote by $x_f$ the smallest solution of $f(x)=x$. We now assume $m\leqslant 1$. Show that $x_f=1$ and that for all $n\in\mathbb{N}$, $u_n\neq 1$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Show that $\lim_{n\rightarrow+\infty}\left(\frac{1}{\varepsilon_{n+1}}-\frac{1}{\varepsilon_n}\right)=\frac{f''(1)}{2}$.
We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$. In this question, we assume $m=1$. We set, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$. Deduce that, as $n$ tends to infinity, $1-u_n\sim\frac{2}{f''(1)n}$. One may use Cesaro's lemma: if $(a_n)_{n\in\mathbb{N}}$ is a sequence of real numbers converging to $l$ and if we set, for $n\in\mathbb{N}^*$, $b_n=\frac{1}{n}(a_1+\cdots+a_n)$, then the sequence $(b_n)_{n\geqslant 1}$ converges to $l$.
Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum. We fix $m$ in $\mathbb{N}^{*}$ and $\varepsilon$ in $\mathbb{R}^{+*}$. Using the Euclidean division of $n$ by $m$, show that there exists an integer $N$ such that for all $n > N$, $$\frac{u_{n}}{n} \geqslant \frac{u_{m}}{m} - \varepsilon$$
Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum. Show $\lim_{n \rightarrow \infty} \frac{u_{n}}{n} = s$.
In this subsection II.A, $a$ is a nonzero real number. We denote by Sol(II.1) the set of complex sequences $\left( z _ { k } \right) _ { k \in \mathbb { N } }$ satisfying the recurrence relation $$\forall k \in \mathbb { N } ^ { * } , \quad z _ { k + 1 } + a z _ { k } + z _ { k - 1 } = 0$$ Suppose in this question that $p$ is an integer greater than or equal to 3. Give a value of $a \in ] - 2,2 [$ for which all solutions of equation (II.1) are $p$-periodic.