Thu-27-03-2025 20:39
profe: Dario Martin Aza
status:
tags: Sucesiones
Ej 7 G1
A A A a ∈ A a \in A a ∈ A b ∈ R b \in \mathbb{R} b ∈ R x 0 = s u p ( A ) x_{0}=sup(A) x 0 = s u p ( A ) x ∈ A x \in A x ∈ A ⟹ f ( x ) > x \implies f(x)>x ⟹ f ( x ) > x x 0 x_{0} x 0 A , A, A ,
x ≤ x 0 ∀ x ∈ A x\leq x_{0}\quad \forall x \in A x ≤ x 0 ∀ x ∈ A Como f f f f ( x 0 ) ≥ f ( x ) > x ∀ x ∈ A f(x_{0})\geq f(x)>x\quad\forall x \in A f ( x 0 ) ≥ f ( x ) > x ∀ x ∈ A f ( x 0 ) f(x_{0}) f ( x 0 ) A A A x 0 = s u p ( A ) ⟹ x 0 ≤ f ( x 0 ) x_{0}=sup(A)\implies x_{0}\leq f(x_{0}) x 0 = s u p ( A ) ⟹ x 0 ≤ f ( x 0 )
Sea ε > 0 \varepsilon>0 ε > 0 x 0 + ε > x 0 x_{0}+\varepsilon>x_{0} x 0 + ε > x 0 f f f
f ( x 0 + ε ) ≥ f ( x 0 ) ≥ ( 1 ) x 0 f(x_{0}+\varepsilon)\geq f(x_{0})\underset{ (1) }{ \geq } x_{0} f ( x 0 + ε ) ≥ f ( x 0 ) ( 1 ) ≥ x 0 Como x 0 + ε > x 0 = s u p ( A ) x_{0}+\varepsilon>x_{0}=sup(A) x 0 + ε > x 0 = s u p ( A ) x 0 + ε ∉ A x_{0}+\varepsilon \not\in A x 0 + ε ∈ A f ( x 0 + ε ) ≤ x 0 + ε f(x_{0}+\varepsilon)\leq x_{0}+\varepsilon f ( x 0 + ε ) ≤ x 0 + ε ⟹ x 0 + ε ≥ f ( x 0 + ε ) ≥ f ( x 0 ) ≥ x 0 ∀ ε > 0 \implies x_{0}+\varepsilon\geq f(x_{0}+\varepsilon)\geq f(x_{0})\geq x_{0}\quad\forall\varepsilon>0 ⟹ x 0 + ε ≥ f ( x 0 + ε ) ≥ f ( x 0 ) ≥ x 0 ∀ ε > 0 x 0 + ε ∈ [ a , b ] x_{0}+\varepsilon \in[a,b] x 0 + ε ∈ [ a , b ] x 0 = b x_{0}=b x 0 = b ε → 0 \varepsilon\to0 ε → 0
x 0 ≥ f ( x 0 ) ≥ x 0 x_{0}\geq f(x_{0})\geq x_{0} x 0 ≥ f ( x 0 ) ≥ x 0 En caso de que x 0 = b x_{0}=b x 0 = b
Para todo n ∈ N n\in \mathbb{N} n ∈ N a n ∈ A a_{n} \in A a n ∈ A b − 1 n < a n ≤ b b- \frac{1}{n}<a_{n}\leq b b − n 1 < a n ≤ b
f ( b ) ≥ f ( a n ) > a n > b − 1 n ∀ n ∈ N f(b)\geq f(a_{n})>a_{n}>b- \frac{1}{n}\quad \forall n \in \mathbb{N} f ( b ) ≥ f ( a n ) > a n > b − n 1 ∀ n ∈ N ⟹ f ( b ) > b − 1 n ∀ n ∈ N \implies f(b)> b- \frac{1}{n} \quad \forall n \in \mathbb{N} ⟹ f ( b ) > b − n 1 ∀ n ∈ N ⟹ f ( b ) ≥ b \implies f(b)\geq b ⟹ f ( b ) ≥ b y como f ( b ) ∈ [ a , b ] : f ( b ) ≤ b f(b)\in[a,b]:f(b)\leq b f ( b ) ∈ [ a , b ] : f ( b ) ≤ b
∴ f ( b ) = b \therefore\: f(b)=b ∴ f ( b ) = b Ejercicio clase pasada Pablo
Probó que entonces l ≥ x n ∀ n ∈ N l\geq x_{n}\quad\forall n \in \mathbb{N} l ≥ x n ∀ n ∈ N ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N ∃ l ^ ∈ R \:\exists\:\hat{l}\in \mathbb{R} ∃ l ^ ∈ R x n → l ^ x_{n}\to \hat{l} x n → l ^ x n k → l ^ x_{n_{k}}\to \hat{l} x n k → l ^ ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N l = l ^ l=\hat{l} l = l ^
Ejercicio 1
Decidir si la sucesión
a n = 1 n 2 + 1 n 2 + 1 + ⋯ + 1 n 2 + n ⏟ n + 1 cosas a_{n}=\underbrace{ \frac{1}{\sqrt{ n^{2} }}+\frac{1}{\sqrt{ n^{2}+1 }}+\dots+\frac{1}{\sqrt{ n^{2}+n }} }_{ \text{n + 1 cosas} } a n = n + 1 cosas n 2 1 + n 2 + 1 1 + ⋯ + n 2 + n 1 es convergente.
D e m : Dem: D e m :
a n ≥ ( n + 1 ) ⋅ 1 n 2 + n a_{n}\geq (n+1)\cdot \frac{1}{\sqrt{ n^{2}+n }} a n ≥ ( n + 1 ) ⋅ n 2 + n 1 a n ≤ ( n + 1 ) ⋅ 1 n a_{n}\leq (n+1)\cdot \frac{1}{n} a n ≤ ( n + 1 ) ⋅ n 1 n + 1 n 2 + n ≤ a n ≤ n + 1 n \frac{n+1}{\sqrt{ n^{2}+n }}\leq a_{n}\leq \frac{n+1}{n} n 2 + n n + 1 ≤ a n ≤ n n + 1 1 ≤ lim n → ∞ a n ≤ 1 1\leq \underset{ n\to \infty }{ \lim }a_{n}\leq 1 1 ≤ n → ∞ lim a n ≤ 1 Por el lema del sandwich a n → 1 a_{n}\to1 a n → 1
Ejercicio 2
Sea ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N
x n + 1 = x n 2 + 2 x n + 1 , x 1 = 5 x_{n+1}=\frac{x_{n}^{2}+2}{x_{n}+1}\quad \quad ,\:x_{1}=5 x n + 1 = x n + 1 x n 2 + 2 , x 1 = 5 Decidir si ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N
D e m : Dem: D e m : x n → L x_{n}\to L x n → L
L = lim n → ∞ x n + 1 = lim n → ∞ x n 2 + 2 x n + 1 = lim n → ∞ L 2 + 2 L + 1 L=\underset{ n\to \infty }{ \lim } x_{n+1}=\underset{ n\to \infty }{ \lim } \frac{x_{n}^{2}+2}{x_{n}+1}=\underset{ n\to \infty }{ \lim } \frac{L^{2}+2}{L+1} L = n → ∞ lim x n + 1 = n → ∞ lim x n + 1 x n 2 + 2 = n → ∞ lim L + 1 L 2 + 2 L = L 2 + 2 L + 1 L 2 + L = L 2 + 2 L = 2 \begin{array}{c}
L=\frac{L^{2}+2}{L+1} \\
\cancel{ L^{2} }+L=\cancel{ L^{2} }+2 \\
L=2
\end{array} L = L + 1 L 2 + 2 L 2 + L = L 2 + 2 L = 2 Afirmo que ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N x n + 1 < x n ∀ n ∈ N x_{n+1}<x_{n}\quad\forall n \in \mathbb{N} x n + 1 < x n ∀ n ∈ N
⟺ x n 2 + 2 < x n + 1 ⋅ ( x n + 1 ) < s i x n + 1 < x n x n ⋅ ( x n + 1 ) = x n 2 + x n ⟺ 2 < x n \iff x_{n}^{2}+2<x_{n+1}\cdot(x_{n}+1)\underset{ si\:x_{n+1}<x_{n} }{ < }x_{n}\cdot(x_{n}+1)=x_{n}^{2}+x_{n}\iff{2}<x_{n} ⟺ x n 2 + 2 < x n + 1 ⋅ ( x n + 1 ) s i x n + 1 < x n < x n ⋅ ( x n + 1 ) = x n 2 + x n ⟺ 2 < x n Luego basta ver que 2 < x n ∀ n ∈ N 2<x_{n}\quad\forall n \in \mathbb{N} 2 < x n ∀ n ∈ N ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N 2 < x n ∀ n ∈ N , 2<x_{n}\quad\forall n\in \mathbb{N}, 2 < x n ∀ n ∈ N , x n > 2 , x_{n}>2, x n > 2 ,
x n + 1 > 2 ⟺ x n 2 + 2 x n + 1 > 2 ⟺ x n 2 + 2 > 2 x n + 2 x_{n+1}>2\iff \frac{x_{n}^{2}+2}{x_{n}+1}>2\iff x_{n}^{2}+2>2x_{n}+2 x n + 1 > 2 ⟺ x n + 1 x n 2 + 2 > 2 ⟺ x n 2 + 2 > 2 x n + 2 ⟺ x n 2 > 2 x n ⟺ x n > 2 \iff x_{n}^{2}>2x_{n}\iff x_{n}>2 ⟺ x n 2 > 2 x n ⟺ x n > 2 Como ( x n ) n ∈ N ( x_{n} )_{n \in \mathbb{N}} ( x n ) n ∈ N x n → 2. x_{n}\to{2}. x n → 2 .
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