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(eroare: eq.0/48690)\frac{C_{4\,n}^{\,2\,n}}{{{4}^{n}}\ C_{2\,n}^{\,n}}\ ,\ \ \ n\in {{\mathbf{N}}^{*}}
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ex. 353
Acum construim alte dou? ?iruri ajut?toare:
(eroare: eq.12/48695)Am ob\c{t}inut $\[{{c}_{n}}<{{a}_{n}}<{{b}_{n}}\ ,\ \ \forall \ n\in {{\mathbf{N}}^{*}}\]$(eroare: eq.13/48695)Observ\u{a}m c\u{a} $\[{{a}_{n}}\cdot {{b}_{n}}=\frac{2\,n+1}{\ 4\,n+1\ }\ \ \xrightarrow[n\ \to \ \infty ]{}\ \ \frac{1}{2}\]$ \c{s}i $\[{{a}_{n}}\cdot {{c}_{n}}=\frac{2\,n}{\ 4\,n\ }\ \ \xrightarrow[n\ \to \ \infty ]{}\ \ \frac{1}{2}\]$(eroare: eq.14/48695)\[\left. {{c}_{n}}<{{a}_{n}}<{{b}_{n}}\ \ \ \right|\ \cdot \ {{a}_{n}}>0\ \ \ \ \ \ \ \ \Leftrightarrow \ \ \ {{a}_{n}}\cdot {{c}_{n}}<a_{n}^{2}<{{a}_{n}}\cdot {{b}_{n}}\](eroare: eq.15/48695)Din teorema cle\c{}stelui $\[\Rightarrow \ \ \exists \ \,\underset{n\ \to \ \infty }{\mathop{\lim }}\,\ a_{n}^{2}=\ \,\frac{1}{2}\ \ \ \Rightarrow \ \ \ \exists \ \,\underset{n\ \to \ \infty }{\mathop{\lim }}\,\ {{a}_{n}}=\ \,\frac{\sqrt{2}}{2}\]$
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(eroare: eq.0/48698)Text oarecare $\[{{c}_{n}}<{{a}_{n}}<{{b}_{n}}\ ,\ \ \forall \ n\in {{\mathbf{N}}^{*}}\]$
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