od Onomatopeja » Sreda, 15. April 2020, 14:34
Sa jedne strane vazi [inlmath]2n^3+(-1)^k\sqrt{k}\geq 2n^3-\sqrt{k}\geq 2n^3-\sqrt{n}>0,[/inlmath] pa je
[dispmath]\frac{\sqrt{n}}{1+\sqrt{2n^3+(-1)^k\sqrt{k}}} \leq \frac{\sqrt{n}}{1+\sqrt{2n^3-\sqrt{n}}},[/dispmath]
odnosno
[dispmath]\sum_{k=0}^n \frac{\sqrt{n}}{1+\sqrt{2n^3+(-1)^k\sqrt{k}}} \leq \sum_{k=0}^n\frac{\sqrt{n}}{1+\sqrt{2n^3-\sqrt{n}}}= \frac{(n+1)\sqrt{n}}{1+\sqrt{2n^3-\sqrt{n}}}.[/dispmath]
Sa druge strane, slicno, vazi [inlmath]0<2n^3+(-1)^k\sqrt{k}\leq 2n^3+\sqrt{k}\leq 2n^3+\sqrt{n},[/inlmath] pa je
[dispmath]\frac{\sqrt{n}}{1+\sqrt{2n^3+(-1)^k\sqrt{k}}} \geq\frac{\sqrt{n}}{1+\sqrt{2n^3+\sqrt{n}}},[/dispmath]
odnosno
[dispmath]\sum_{k=0}^n \frac{\sqrt{n}}{1+\sqrt{2n^3+(-1)^k\sqrt{k}}} \geq \sum_{k=0}^n\frac{\sqrt{n}}{1+\sqrt{2n^3+\sqrt{n}}}= \frac{(n+1)\sqrt{n}}{1+\sqrt{2n^3+\sqrt{n}}}.[/dispmath]
Dakle, za svako [inlmath]n\in\mathbb{N}[/inlmath] vazi
[dispmath]\frac{(n+1)\sqrt{n}}{1+\sqrt{2n^3+\sqrt{n}}}\leq\sum_{k=0}^n \frac{\sqrt{n}}{1+\sqrt{2n^3+(-1)^k\sqrt{k}}} \leq \frac{(n+1)\sqrt{n}}{1+\sqrt{2n^3-\sqrt{n}}}.[/dispmath]
Kako je [inlmath]\displaystyle\lim_{n\to\infty} \frac{(n+1)\sqrt{n}}{1+\sqrt{2n^3+\sqrt{n}}}=\frac1{\sqrt{2}}[/inlmath] i [inlmath]\displaystyle\lim_{n\to\infty} \frac{(n+1)\sqrt{n}}{1+\sqrt{2n^3-\sqrt{n}}}=\frac1{\sqrt{2}},[/inlmath] to je prema lemi o dva policajca
[dispmath]\lim_{n\to\infty} \sum_{k=0}^n \frac{\sqrt{n}}{1+\sqrt{2n^3+(-1)^k\sqrt{k}}} = \frac{1}{\sqrt{2}}.[/dispmath]