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Δημοσιεύτηκε: **Σάβ Αύγ 13, 2011 10:51 pm**

Έστω η συνάρτηση $f:\mathbb{R}\rightarrow\mathbb{R}$ που είναι τέτοια ώστε να ισχύουν:
$\displaystyle\ f(1)=11$
$\displaystyle\ f(x+1)=\frac{(f(x))^{2}+f(x)-1}{f(x)}$
Εάν $\displaystyle\ A=\sum_{i=1}^{n}\frac{1}{f(i)-1}, B=\sum_{i=1}^{n}\frac{1}{f(i)+1}$ , τότε να υπολογίσετε τα όρια
1) $\dislaystyle\\lim_{n\rightarrow\infty }(A+B)$
2) $\displaystyle\\lim_{n\rightarrow\infty }(A-B)$

Δημοσιεύτηκε: **Πέμ Αύγ 25, 2011 8:16 pm**

Θα αποδείξουμε διαδοχικά :

1) $\displaystyle{f\left( n \right) > 1}$

2) H ακολουθία f(n)

είναι γνησίως αύξουσα.

3) $\displaystyle{\mathop {\lim }\limits_{n \to \infty } f\left( n \right) = + \infty }$

4) $\displaystyle{{A = B + \frac{1}{5} - \frac{2}{{f\left( {n + 1} \right) - 1}}}}$

5) $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \frac{{f\left( n \right)}}{n} = 1}$

6) $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {{\rm A} - {\rm B}} \right) = \frac{1}{5}}$

7) $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {A + B} \right) = + \infty }$

Ισχύει $\displaystyle{f\left( {n + 1} \right) = \frac{{{f^2}\left( n \right) + f\left( n \right) - 1}}{{f\left( n \right)}} \Rightarrow f\left( {n + 1} \right) = f\left( n \right) + 1 - \frac{1}{{f\left( n \right)}}}$

1) $\displaystyle{f\left( 1 \right) = 11 > 1}$ , Έστω $\displaystyle{f\left( n \right) > 1 \Rightarrow \frac{1}{{f\left( n \right)}} < 1 \Rightarrow - \frac{1}{{f\left( n \right)}} > - 1 \Rightarrow 1 + f\left( n \right) - \frac{1}{{f\left( n \right)}} > 1 + f\left( n \right) - 1 > 1 \Rightarrow f\left( {n + 1} \right) > 1}$ .

2) $\displaystyle{f\left( {n + 1} \right) = f\left( n \right) + 1 - \frac{1}{{f\left( n \right)}} \Rightarrow f\left( {n + 1} \right) - f\left( n \right) = 1 - \frac{1}{{f\left( n \right)}} > 0 \Rightarrow f\left( n \right) \uparrow }$ , δηλαδή τελικά f(n)>11

.

3) Η

είναι γνησίως αύξουσα, επομένως είτε θα συγκλίνει, είτε θα απειρίζεται. Έστω $\displaystyle{\mathop {\lim }\limits_{n \to \infty } f\left( n \right) = \lambda }$ , τότε $\displaystyle{f\left( {n + 1} \right) = f\left( n \right) + 1 - \frac{1}{{f\left( n \right)}} \Rightarrow }$
$\displaystyle{ \Rightarrow \mathop {\lim }\limits_{n \to \infty } f\left( {n + 1} \right) = \mathop {\lim }\limits_{n \to \infty } f\left( n \right) + 1 - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{f\left( n \right)}} \Rightarrow \lambda = \lambda + 1 - \frac{1}{\lambda } \Rightarrow \lambda = 1}$ άτοπο διότι f(n)>11

, άρα $\displaystyle{\mathop {\lim }\limits_{n \to \infty } f\left( n \right) = + \infty }$ .

4) $\displaystyle{f\left( {k + 1} \right) = \frac{{{f^2}\left( k \right) + f\left( k \right) - 1}}{{f\left( k \right)}} \Rightarrow f\left( {k + 1} \right) - 1 = \frac{{{f^2}\left( k \right) - 1}}{{f\left( k \right)}} \Rightarrow \frac{1}{{f\left( {k + 1} \right) - 1}} = \frac{1}{2}\left( {\frac{1}{{f\left( k \right) - 1}} + \frac{1}{{f\left( k \right) + 1}}} \right) \Rightarrow }$

$\displaystyle{ \Rightarrow 2\sum\limits_{k = 1}^n {\frac{1}{{f\left( {k + 1} \right) - 1}}} = \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) - 1}}} + \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) + 1}}} \Rightarrow 2\sum\limits_{k = 2}^{n + 1} {\frac{1}{{f\left( k \right) - 1}}} = \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) - 1}}} + \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) + 1}}} \Rightarrow }$

$\displaystyle{ \Rightarrow 2\left( { - \frac{1}{{f\left( 1 \right) - 1}} + \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) - 1}}} + \frac{1}{{f\left( {n + 1} \right) - 1}}} \right) = \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) - 1}}} + \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) + 1}}} \Rightarrow }$

$\displaystyle{ \Rightarrow - \frac{1}{5} + 2A + \frac{2}{{f\left( {n + 1} \right) - 1}} = A + B \Rightarrow \boxed{A = B + \frac{1}{5} - \frac{2}{{f\left( {n + 1} \right) - 1}}}}$ .

5) $\displaystyle{f\left( {n + 1} \right) = f\left( n \right) + 1 - \frac{1}{{f\left( n \right)}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( {n + 1} \right) - f\left( n \right)}}{{\left( {n + 1} \right) - n}} = 1 - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{f\left( n \right)}} = 1 \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( n \right)}}{n} = 1}$ (Stolz - Secaro).

6) $\displaystyle{A = B + \frac{1}{5} - \frac{2}{{f\left( {n + 1} \right) - 1}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {{\rm A} - {\rm B}} \right) = \frac{1}{5} - \mathop {\lim }\limits_{n \to \infty } \frac{2}{{f\left( {n + 1} \right) - 1}} = \frac{1}{5}}$

7) $\displaystyle{f\left( n \right) = \varphi \left( n \right) \cdot n{\text{ }}{\text{, }}\mathop {\lim }\limits_{n \to \infty } \frac{{f\left( n \right)}}{n} = 1 \Rightarrow \mathop {\lim }\limits_{n \to \infty } \varphi \left( n \right) = 1 \Rightarrow \forall \varepsilon > 0:\exists {n_0} \in N:\forall n \geqslant {n_0}:\varphi \left( n \right) < 1 + \varepsilon }$

επομένως $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{f\left( k \right) - 1}}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k \cdot \varphi \left( k \right) - 1}}} = \sum\limits_{k = 1}^{{n_0} - 1} {\frac{1}{{k \cdot \varphi \left( k \right) - 1}}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = {n_0}}^n {\frac{1}{{k \cdot \varphi \left( k \right) - 1}}} > M + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = {n_0}}^n {\frac{1}{{k \cdot \left( {1 + \varepsilon } \right) - 1}}} = }$

$\displaystyle{ = M + \frac{1}{{1 + \varepsilon }} \cdot \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = {n_0}}^n {\frac{1}{{k - 1/\left( {1 + \varepsilon } \right)}}} > M + \frac{1}{{1 + \varepsilon }} \cdot \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = {n_0}}^n {\frac{1}{{k - 1}}} = + \infty }$ . (Αρμονική σειρά)

Άρα $\displaystyle{\mathop {\lim }\limits_{n \to \infty } A = \mathop {\lim }\limits_{n \to \infty } B = \mathop {\lim }\limits_{n \to \infty } \left( {A + B} \right) = + \infty }$ .

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