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Δημοσιεύτηκε: **Δευ Απρ 30, 2012 4:32 pm**

Έστω

ακολουθία θετικών αριθμών με $\displaystyle{ \lim_{n \to \infty} x_n = 0.}$ Να υπολογιστεί το $\displaystyle{ \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \ln{\left(\frac{k}{n}+x_n \right).}}$

Δημοσιεύτηκε: **Σάβ Ιούλ 28, 2012 9:04 pm**

Demetres έγραψε:Έστω ακολουθία θετικών αριθμών με $\displaystyle{ \lim_{n \to \infty} x_n = 0.}$ Να υπολογιστεί το $\displaystyle{ \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \ln{\left(\frac{k}{n}+x_n \right).}}$

$\displaystyle{\left\{ {{x_n}} \right\}:}$ ακολουθία θετικών όρων και $\displaystyle{\mathop {\lim }\limits_{n \to \infty } {x_n} = 0}$ . Τότε $\displaystyle{\forall \varepsilon > 0:\exists {n_0}:\forall n > {n_0}:0 < {x_n} < \varepsilon }$ . Επομένως αν $\displaystyle{m\;,\;M}$ τα κάτω και άνω φράγματα

της $\displaystyle{{\left\{ {{x_n}} \right\}_{n = 1\,,\,2\,,\,..\,,\,{n_0}}}}$ θα ισχύει $\displaystyle{\sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + m} \right)} + \sum\limits_{k = {n_0} + 1}^n {\ln \left( {\frac{k}{n}} \right)} \leqslant \sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n} + {x_n}} \right)} \leqslant \sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + M} \right)} + \sum\limits_{k = {n_0} + 1}^n {\ln \left( {\frac{k}{n} + \varepsilon } \right)} }$ ,

δηλαδή $\displaystyle{\sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + m} \right)} + \sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n}} \right)} - \sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n}} \right)} \leqslant \sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n} + {x_n}} \right)} \leqslant \sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + M} \right)} + \sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n} + \varepsilon } \right)} - \sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + \varepsilon } \right)} }$

Όμως $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n}} \right)} } \right) = \int\limits_0^1 {\ln \left( x \right)dx} = - 1}$ και $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n} + \varepsilon } \right)} } \right) = \int\limits_0^1 {\ln \left( {x + \varepsilon } \right)dx} = - 1 - \varepsilon \ln \varepsilon + \left( {1 + \varepsilon } \right)\ln \left( {1 + \varepsilon } \right)\xrightarrow{{\varepsilon \to 0}} - 1}$
(ολοκληρώματα Riemann)

Τα όρια $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}\sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + m} \right)} } \right)\;,\quad \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}\sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + M} \right)} } \right)\;\& \quad \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}\sum\limits_{k = 1}^{{n_0}} {\ln \left( {\frac{k}{n} + \varepsilon } \right)} } \right)}$ , είναι όλα μηδέν (σχεδόν προφανές).

Επίσης $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^{{n_0}} {\ln \frac{k}{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\ln \left( {\prod\limits_{k = 1}^{{n_0}} {\frac{k}{n}} } \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\ln \left( {\frac{{{n_0}!}}{{{n^{{n_0}}}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \ln \left( {\frac{{{{\left( {{n_0}!} \right)}^{1/n}}}}{{{{\left( {{n^{1/n}}} \right)}^{{n_0}}}}}} \right) = \ln 1 = 0}$ .

Από τα παραπάνω συνάγεται ότι $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\ln \left( {\frac{k}{n} + {x_n}} \right)} } \right) = - 1}$ .

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IMC 1997/1/1

IMC 1997/1/1

Re: IMC 1997/1/1