Foro de estudiantes

Sean las sucesiones $$b_{n}$$ y $$c_{n}$$

$$\left\lbrace \begin{array}{ccl} b_{2} & = & a_{0} \\ b_{n+1} & = & \left\lbrace \begin{array}{l} a_{n} \textup{ si } a_{n}\leq \sqrt{2} \\ b_{n} \textup{ si } a_{n}>\sqrt{2} \end{array} \forall n\geq 2 \right. \end{array} \right.$$

$$\left\lbrace \begin{array}{ccl} c_{2} & = & a_{1} \\ c_{n+1} & = & \left\lbrace \begin{array}{l} c_{n} \textup{ si } a_{n}\leq \sqrt{2} \\ a_{n} \textup{ si } a_{n}>\sqrt{2} \end{array} \forall n\geq 2 \right. \end{array} \right.$$

Se demuestra por inducción que $$c_{n}-a_{n}=a_{n}-b_{n}=\frac{a_{1}-a_{0}}{2^{n-1}}\hspace{0.25cm} \forall n\geq 2$$ y que $$b_{n}\leq \sqrt{2}\leq c_{n}\hspace{0.25cm} \forall n\geq 2$$.

Paso base:

$$c_{2}-a_{2}=a_{1}-\left(a_{1}-\frac{a_{1}-a_{0}}{2}\right)=\frac{a_{1}-a_{0}}{2}$$

$$a_{2}-b_{2}=a_{1}-\frac{a_{1}-a_{0}}{2}-a_{0}=\frac{a_{1}-a_{0}}{2}$$

$$b_{2}=a_{0}\leq \sqrt{2}\leq a_{1}=c_{2}\Rightarrow b_{2}\leq \sqrt{2}\leq c_{2}$$

Paso inductivo:

Supongamos que $$c_{k}-a_{k}=a_{k}-b_{k}=\frac{a_{1}-a_{0}}{2^{k-1}}$$ y que $$b_{k}\leq \sqrt{2}\leq c_{k}$$ para algún $$k\geq 2$$.

Caso 1: si $$a_{k}\leq \sqrt{2}$$

Entonces $$b_{k+1}=a_{k}$$, $$c_{k+1}=c_{k}$$ y $$a_{k+1}=a_{k}+\frac{a_{1}-a_{0}}{2^{k}}$$.

$$c_{k+1}-a_{k+1}=c_{k}-a_{k}-\frac{a_{1}-a_{0}}{2^{k}}=\frac{a_{1}-a_{0}}{2^{k-1}}-\frac{a_{1}-a_{0}}{2^{k}}=\frac{a_{1}-a_{0}}{2^{k}}$$

$$a_{k+1}-b_{k+1}=a_{k}+\frac{a_{1}-a_{0}}{2^{k}}-a_{k}=\frac{a_{1}-a_{0}}{2^{k}}$$

$$b_{k+1}=a_{k}\leq \sqrt{2}\Rightarrow b_{k+1}\leq \sqrt{2}$$

$$c_{k+1}=c_{k}\geq \sqrt{2}\Rightarrow c_{k+1}\geq \sqrt{2}$$

Caso 2: si $$a_{k}> \sqrt{2}$$

Entonces $$b_{k+1}=b_{k}$$, $$c_{k+1}=a_{k}$$ y $$a_{k+1}=a_{k}-\frac{a_{1}-a_{0}}{2^{k}}$$.

$$c_{k+1}-a_{k+1}=a_{k}-a_{k}+\frac{a_{1}-a_{0}}{2^{k}}=\frac{a_{1}-a_{0}}{2^{k}}$$

$$a_{k+1}-b_{k+1}=a_{k}-\frac{a_{1}-a_{0}}{2^{k}}-b_{k}=\frac{a_{1}-a_{0}}{2^{k-1}}-\frac{a_{1}-a_{0}}{2^{k}}=\frac{a_{1}-a_{0}}{2^{k}}$$

$$b_{k+1}=b_{k}\leq \sqrt{2}\Rightarrow b_{k+1}\leq \sqrt{2}$$

$$c_{k+1}=a_{k}> \sqrt{2}\Rightarrow c_{k+1}\geq \sqrt{2}$$

En cualquier caso, se cumple el paso inductivo.

Entonces, $$b_{n}\leq \sqrt{2}\leq c_{n} \hspace{0.25cm} \forall n\geq 2$$,    $$b_{n}\leq a_{n}\leq c_{n}\hspace{0.25cm} \forall n\geq 2$$, y $$\left|c_{n}-b_{n}\right|=\frac{a_{1}-a_{0}}{2^{n-2}}\hspace{0.25cm} \forall n\geq 2$$.

Por lo tanto, $$|a_{n}-\sqrt{2}|\leq |c_{n}-b_{n}| \Rightarrow |a_{n}-\sqrt{2}|\leq \frac{a_{1}-a_{0}}{2^{n-2}}\hspace{0.25cm} \forall n\geq 2$$

Como $$\lim_{n\to +\infty}\frac{a_{1}-a_{0}}{2^{n-2}}=0$$, entonces $$\forall \varepsilon>0, \exists n_{0} : \frac{a_{1}-a_{0}}{2^{n-2}}<\varepsilon \hspace{0.25cm} \forall n\geq n_{0}$$

$$\forall \varepsilon >0, \exists n_{0} : |a_{n}-\sqrt{2}|\leq \varepsilon \hspace{0.25cm} \forall n\geq n_{0}$$

Entonces $$\lim_{n\to +\infty} |a_{n}-\sqrt{2}|=0\Rightarrow \lim_{n\to +\infty} a_{n}=\sqrt{2}$$.

La parte b) es fácil sabiendo que $$|a_{n}-\sqrt{2}|\leq \frac{a_{1}-a_{0}}{2^{n-2}} \hspace{0.25cm} \forall n\geq 2$$

ya que simplemente se despeja $$n$$:

$$\frac{a_{1}-a_{0}}{2^{n-2}}<\varepsilon \Rightarrow 2^{n-2}>\frac{a_{1}-a_{0}}{\varepsilon} \Rightarrow n-2>\log_{2} \left( \frac{a_{1}-a_{0}}{\varepsilon} \right) \Rightarrow n>\log_{2} \left( \frac{a_{1}-a_{0}}{\varepsilon} \right)+2$$