Rozwiązanie:
Ciąg:
[tex]$a_{n}=\sqrt{(n+a)(n+b)} -n[/tex]
Granica:
[tex]$\lim_{n \to \infty} a_{n}=\lim_{n \to \infty}\sqrt{(n+a)(n+b)}-n =[/tex]
[tex]$=\lim_{n \to \infty} \frac{(\sqrt{(n+a)(n+b)}-n )(\sqrt{(n+a)(n+b)}+n )}{\sqrt{(n+a)(n+b)}+n } =[/tex]
[tex]$= \lim_{n \to \infty} \frac{(n+a)(n+b)-n^{2}}{\sqrt{(n+a)(n+b)}+n } = \lim_{n \to \infty}\frac{n^{2}+bn+an+ab-n^{2}}{\sqrt{(n+a)(n+b)}+n} =[/tex]
[tex]$ \lim_{n \to \infty} \frac{bn+an+ab}{\sqrt{(n+a)(n+b)}+n} = \lim_{n \to \infty} \frac{n\Big(a+b+\frac{ab}{n} \Big)}{n\Big(\sqrt{1+\frac{b}{n}+\frac{a}{n}+\frac{ab}{n^{2}} }+1 \Big)} =[/tex]
[tex]$\lim_{n \to \infty} \frac{a+b+\frac{ab}{n}}{\sqrt{1+\frac{b}{n}+\frac{a}{n}+\frac{ab}{n^{2}} }+1 }=\frac{a+b+0}{\sqrt{1+0+0+0} +1} =\frac{a+b}{2}[/tex]
co kończy dowód.
