Rozwiązanie:
[tex]$ \lim_{n \to \infty} \frac{{n \choose 1} {n \choose 2} {n \choose 3}}{(1+2+...+n)^{2}}[/tex]
Licznik:
[tex]${n \choose 1} {n \choose 2} {n \choose 3}=\frac{n!}{1!(n-1)!} \cdot \frac{n!}{2!(n-2)!} \cdot \frac{n!}{3!(n-3)!} =[/tex]
[tex]$=n \cdot \frac{n(n-1)}{2} \cdot \frac{n(n-1)(n-2)}{6}=\frac{n^{3}(n-1)^{2}(n-2)}{12}[/tex]
Mianownik:
[tex]$1+2+3+...+n= \sum\limits^{n}_{k=1}k=\frac{n(n+1)}{2}[/tex]
Wstawiamy do granicy:
[tex]$ \lim_{n \to \infty} \frac{{n \choose 1} {n \choose 2} {n \choose 3}}{(1+2+...+n)^{2}}= \lim_{n \to \infty} \frac{n^{3}(n-1)^{2}(n-2)}{12} \cdot \frac{2}{n(n+1)} =[/tex]
[tex]$= \lim_{n \to \infty} \frac{n^{2}(n-1)^{2}(n-2)}{6(n+1)} = \lim_{n \to \infty} \frac{n^{5}\Big(1-\frac{1}{n} \Big)^{2}\Big(1-\frac{2}{n} \Big)}{n\Big(6+\frac{1}{n} \Big)} =\infty[/tex]
