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Question Number 205672 by hardmath last updated on 26/Mar/24

Answered by Berbere last updated on 27/Mar/24

$$\mathrm{\Omega }=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=2}{\sum ^n}\frac{\sqrt{k(k-1)}}{\sqrt{{(\sqrt{k-1}+\sqrt{k})}^2}}.{\left(\underset{k=1}{\sum ^n}\sqrt{k}\right)}^{-1}$$$$S_1=\underset{k=2}{\sum ^n}\frac{\sqrt{k(k-1)}}{\sqrt{k-1}+k}=\underset{k=1}{\sum ^n}\frac{\sqrt{k(k-1)}}{\sqrt{k-1}+\sqrt{k}}=\underset{k=2}{\sum ^n}(k\sqrt{(k-1)}-(k-1)\sqrt{k})$$$$=\underset{k=2}{\sum ^n}((k-1)\sqrt{k-1}+\sqrt{k-1}-k\sqrt{k}+\sqrt{k)}$$$$=\underset{k=2}{\sum ^n}((k-1)\sqrt{k-1}-k\sqrt{k})+\underset{k=2}{\sum ^n}(\sqrt{k}+\sqrt{k-1})$$$$=-n\sqrt{n}+1+(2\underset{k=2}{\sum ^n}\sqrt{k}-\sqrt{n-1}+1)$$$$\sqrt{k-1}⩽{\int }_{k-1}^k\sqrt{x}⩽\sqrt{k}$$$$\underset{k=2}{\sum ^n}\sqrt{k-1}⩽{\int }_1^n\sqrt{x}⩽\underset{k=2}{\sum ^n}\sqrt{k}=S$$$$S-\sqrt{n}+1⩽\frac{2}{3}n\sqrt{n}-\frac{2}{3}⩽S$$$$S\sim \frac{2}{3}n\sqrt{n}$$$$\mathrm{\Omega }=\frac{1}{n}(-n\sqrt{n}+2+2S-2\sqrt{n-1+})/S$$$$\mathrm{\Omega }\sim \frac{1}{n}\left(\frac{1}{3}n\sqrt{n}\right).{\left(\frac{2}{3}n\sqrt{n}\right)}^{-1}=0$$$$$$$$$$$$$$$$$$

Commented by hardmath last updated on 27/Mar/24

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathrm{professor},\mathrm{perfect}$$

Commented by Berbere last updated on 27/Mar/24

$$WithePleasur$$

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