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Question Number 204472 by mnjuly1970 last updated on 18/Feb/24

$$$$$$\mathrm{Calculate}...$$$$\mathrm{\Omega }=\underset{k=1}{\sum ^n}\lfloor \frac{1}{\sqrt[k]{e}-1}\rfloor =?$$$$$$

Answered by TonyCWX08 last updated on 18/Feb/24

$$whenk=1$$$$\lfloor \frac{1}{e-1}\rfloor =\lfloor 0.582\rfloor =0$$$$$$$$whenk=2$$$$\lfloor \frac{1}{\sqrt{e}-1}\rfloor =\lfloor 1.541\rfloor =1$$$$$$$$whenk=3$$$$\lfloor \frac{1}{e^{\frac{1}{3}}-1}\rfloor =\lfloor 2.528\rfloor =2$$$$$$$$whenk=4$$$$\lfloor \frac{1}{e^{\frac{1}{4}}-1}\rfloor =\lfloor 3.521\rfloor =3$$$$$$$$So$$$$\underset{k=1}{\sum ^n}\lfloor \frac{1}{\sqrt[k]{e}-1}\rfloor$$$$=\underset{k=1}{\sum ^n}(k-1)$$$$=\frac{n^2}{2}$$

Answered by witcher3 last updated on 18/Feb/24

$${\mathrm{e}}^{\frac{1}{\mathrm{k}}}$$$$1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}\mathrm{e}⩾{\mathrm{e}}^{\mathrm{x}}>1+\mathrm{x}$$$$\Rightarrow \frac{1}{\mathrm{k}}+\frac{\mathrm{e}}{{2\mathrm{k}}^2}⩾{\mathrm{e}}^{\frac{1}{\mathrm{k}}}-1<\frac{1}{\mathrm{k}}$$$$\Rightarrow \frac{{2\mathrm{k}}^2}{2\mathrm{k}+\mathrm{e}}⩽\frac{1}{{\mathrm{e}}^{\frac{1}{\mathrm{k}}}-1}⩽\mathrm{k}$$$$\Rightarrow \mathrm{k}-1<\mathrm{k}-\frac{\mathrm{e}}{2\mathrm{k}+\mathrm{e}}⩽\frac{1}{{\mathrm{e}}^{\frac{1}{\mathrm{k}}}-1}<\mathrm{k}$$$$\Rightarrow \left[\frac{1}{{\mathrm{e}}^{\frac{1}{\mathrm{k}}}-1}\right]=\mathrm{k}-1$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\mathrm{k}-1=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$$

Commented by TonyCWX08 last updated on 19/Feb/24

$$$$$$theansweriswrong.$$$$\underset{k=1}{\sum ^n}k=\frac{n(n-1)}{2}$$$$But$$$$\underset{k=1}{\sum ^n}k-1\ne \frac{n(n-1)}{2}$$$$itshouldbeequalto\frac{n^2}{2}$$$$$$

Commented by witcher3 last updated on 19/Feb/24

$$\frac{{\mathrm{n}}^2}{2}\notin \mathbb{N}\mathrm{for}\mathrm{n}=2\mathrm{k}+1$$$$\underset{1}{\sum ^{\mathrm{n}}}\mathrm{k}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(\mathrm{k}+\mathrm{a})=\frac{\mathrm{n}(\mathrm{n}+2\mathrm{a}+1))}{2}$$

Commented by mr W last updated on 19/Feb/24

$$itwasmeant:$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(\mathrm{k}-1)=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$$

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