Ένα όριο με γινόμενο και άθροισμα (2)

#1

Ας βρεθεί, αν υπάρχει, το όριο $\displaystyle{\lim_{n\to+\infty}\frac{1}{\sqrt{n}}\left(1+\sum_{k=2}^{n}\prod_{i=1}^{k-1}\frac{n-i}{n+(i+1)}\right)}$

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

#2

$\displaystyle{1 + \sum\limits_{k = 2}^n {\prod\limits_{i = 1}^{k - 1} {\frac{{n - i}}{{n + \left( {i + 1} \right)}}} } =1+ \left[ {\frac{{n - 1}}{{n + 2}} + \frac{{\left( {n - 1} \right) \cdot \left( {n - 2} \right)}}{{\left( {n + 2} \right) \cdot \left( {n + 3} \right)}} + {\text{ }}..{\text{ }} + \frac{{\left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot {\text{ }}..{\text{ }} \cdot 2 \cdot 1}}{{\left( {n + 2} \right) \cdot \left( {n + 3} \right) \cdot {\text{ }}..{\text{ }} \cdot \left( {2 \cdot n} \right)}}} \right] = }$

$\displaystyle{ = 1 + \frac{1}{n} \cdot \left[ {\frac{{n \cdot \left( {n - 1} \right)}}{{n + 2}} + \frac{{n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right)}}{{\left( {n + 2} \right) \cdot \left( {n + 3} \right)}} + {\text{ }}..{\text{ }} + \frac{{n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot {\text{ }}..{\text{ }} \cdot 2 \cdot 1}}{{\left( {n + 2} \right) \cdot \left( {n + 3} \right) \cdot {\text{ }}..{\text{ }} \cdot \left( {2 \cdot n} \right)}}} \right] = }$

$\displaystyle{ = \frac{n}{n} + \frac{1}{n} \cdot \left[ {\frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {n - 2} \right)! \cdot \left( {n + 2} \right)!}} + \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {n - 3} \right)! \cdot \left( {n + 3} \right)!}} + \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {n - 4} \right)! \cdot \left( {n + 4} \right)!}} + {\text{ }}..{\text{ }} + \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {n - n} \right)! \cdot \left( {n + n} \right)!}}} \right] = }$

$\displaystyle{ = \frac{n}{n} + \frac{1}{n} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \left[ {\frac{{\left( {2 \cdot n} \right)!}}{{\left( {n - 2} \right)! \cdot \left( {n + 2} \right)!}} + \frac{{\left( {2 \cdot n} \right)!}}{{\left( {n - 3} \right)! \cdot \left( {n + 3} \right)!}} + \frac{{\left( {2 \cdot n} \right)!}}{{\left( {n - 4} \right)! \cdot \left( {n + 4} \right)!}} + {\text{ }}..{\text{ }} + \frac{{\left( {2 \cdot n} \right)!}}{{\left( {n - n} \right)! \cdot \left( {n + n} \right)!}}} \right] = }$

$\displaystyle{ = \frac{1}{n} \cdot \left[ {\frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 1} \\ \end{array} } \right)} \right] + \frac{1}{n} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \left[ {\left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 2} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 3} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 4} \\ \end{array} } \right) + {\text{ }}..{\text{ }} + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ 0 \\ \end{array} } \right)} \right] = }$

$\displaystyle{ = \frac{1}{n} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \left[ {\left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 2} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 3} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 4} \\ \end{array} } \right) + {\text{ }}..{\text{ }} + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ 0 \\ \end{array} } \right)} \right] = beacause\left\{ {\left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - m} \\ \end{array} } \right) = \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n + m} \\ \end{array} } \right)} \right\} = }$

$\displaystyle{ = \frac{1}{{2 \cdot n}} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \left[ {\left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 2} \\ \end{array} } \right) + {\text{ }}..{\text{ }} + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ 0 \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n + 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n + 2} \\ \end{array} } \right) + {\text{ }}..{\text{ }} + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {2 \cdot n} \\ \end{array} } \right)} \right] = }$

$\displaystyle{ = \frac{1}{{2 \cdot n}} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \left[ {\left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n - 2} \\ \end{array} } \right) + {\text{ }}..{\text{ }} + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ 0 \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n + 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {n + 2} \\ \end{array} } \right) + {\text{ }}..{\text{ }} + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ {2 \cdot n} \\ \end{array} } \right) + \left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ n \\ \end{array} } \right)} \right] - \frac{{n + 1}}{{2 \cdot n}} = }$

$\displaystyle{ = \frac{1}{{2 \cdot n}} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot \sum\limits_{k = 0}^{2 \cdot n} {\left( {\begin{array}{*{20}{c}} {2 \cdot n} \\ k \\ \end{array} } \right)} - \frac{{n + 1}}{{2 \cdot n}} \Rightarrow \boxed{1 + \sum\limits_{k = 2}^n {\prod\limits_{i = 1}^{k - 1} {\frac{{n - i}}{{n + \left( {i + 1} \right)}}} } = \frac{1}{{2 \cdot n}} \cdot \frac{{n! \cdot \left( {n + 1} \right)!}}{{\left( {2 \cdot n} \right)!}} \cdot {2^{2 \cdot n}} - \frac{{n + 1}}{{2 \cdot n}}}}$

Με χρήση του ασυμπτωτικού τύπου του Stirling, δηλαδή $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{\sqrt {2 \cdot \pi \cdot n} }} \cdot {\left( {\frac{e}{n}} \right)^n} = 1}$ έχουμε

$\displaystyle{\frac{1}{{\sqrt n }} \cdot \left( {1 + \sum\limits_{k = 2}^n {\prod\limits_{i = 1}^{k - 1} {\frac{{n - i}}{{n + \left( {i + 1} \right)}}} } } \right) \approx \frac{1}{{\sqrt n }} \cdot \frac{1}{{2 \cdot n}} \cdot \frac{{\sqrt {2 \cdot \pi \cdot n} \cdot {{\left( {\dfrac{n}{e}} \right)}^n} \cdot \sqrt {2 \cdot \pi \cdot \left( {n + 1} \right)} \cdot {{\left( {\dfrac{{n + 1}}{e}} \right)}^{n + 1}}}}{{\sqrt {2 \cdot \pi \cdot \left( {2 \cdot n} \right)} \cdot {{\left( {\dfrac{{2 \cdot n}}{e}} \right)}^{2 \cdot n}}}} \cdot {2^{2 \cdot n}} - \frac{{n + 1}}{{2 \cdot n \cdot \sqrt n }}}$

Αγνοώντας τον παράγοντα $\displaystyle{ - \frac{{n + 1}}{{2 \cdot n \cdot \sqrt n }}}$ που έχει όριο μηδέν έχουμε

$\displaystyle{\frac{1}{{\sqrt n }} \cdot \left( {1 + \sum\limits_{k = 2}^n {\prod\limits_{i = 1}^{k - 1} {\frac{{n - i}}{{n + \left( {i + 1} \right)}}} } } \right) \approx \frac{{\sqrt {2 \cdot \pi \cdot \left( {n + 1} \right)} }}{{\sqrt {\left( {2 \cdot n} \right)} }} = \frac{1}{2} \cdot \frac{{n + 1}}{n} \cdot \frac{{\sqrt \pi }}{e} \cdot {\left( {\frac{{n + 1}}{n}} \right)^n} \cdot \sqrt {\frac{{n + 1}}{n}} }$

και με δεδομένο ότι $\displaystyle{\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{n + 1}}{n}} \right)^n} = e}$ , προκύπτει $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt n }} \cdot \left( {1 + \sum\limits_{k = 2}^n {\prod\limits_{i = 1}^{k - 1} {\frac{{n - i}}{{n + \left( {i + 1} \right)}}} } } \right) = \frac{{\sqrt \pi }}{2}}$

Ένα όριο με γινόμενο και άθροισμα (2)

Ένα όριο με γινόμενο και άθροισμα (2)

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