Rozwiązanie:
[tex]a.[/tex]
[tex]$a_{n}=(1-\frac{1}{2^{2}} )(1-\frac{1}{3^{2}} )(1-\frac{1}{4^{2}} ) \cdot ... \cdot (1-\frac{1}{n^{2}})[/tex]
[tex]$a_{n}=\prod\limits^{n}_{k=2} (1-\frac{1}{k^{2}} )=\prod\limits^{n}_{k=2}(\frac{k^{2}-1}{k^{2}} )=\prod\limits^{n}_{k=2}\frac{(k-1)(k+1)}{k^{2}} =[/tex]
[tex]$=\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2 \cdot 4}{3 \cdot 3} \cdot \frac{3 \cdot 5}{5 \cdot 5 } ... \cdot \frac{(n-1)(n+1)}{n \cdot n} =\frac{1}{2} (1+\frac{1}{n} )=\frac{n+1}{2n}[/tex]
Zatem:
[tex]$ \lim_{n \to \infty} a_n =\lim_{n \to \infty} \frac{n+1}{2n} =\frac{1}{2}[/tex]
[tex]b[/tex].
[tex]$b_{n}=\frac{2^{n}}{n!}[/tex]
Zauważmy, że:
[tex]$\bigwedge \limits^{}_{n \in \mathbb{N}} 2^{n}>0 \wedge \bigwedge \limits^{}_{n \in \mathbb{N}} n!>0 \Rightarrow \bigwedge \limits^{}_{n \in \mathbb{N}} \frac{2^{n}}{n!} >0[/tex]
[tex]$\frac{2^{n}}{n!} =\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot ... \cdot 2 \cdot 2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot (n-1) \cdot n} <\frac{4}{n}[/tex]
Zatem mamy:
[tex]$0< \frac{2^{n}}{n!} <\frac{4}{n}[/tex]
[tex]$0<\lim_{n \to \infty} \frac{2^{n}}{n!} <\lim_{n \to \infty} \frac{4}{n} \Rightarrow \lim_{n \to \infty} \frac{2^{n}}{n!} =0[/tex]
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