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Question Number 213511 by universe last updated on 07/Nov/24

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\frac{1}{2^{\mathrm{r}}}\right]$$$$\mathrm{where}[\bullet ]\mathrm{greatest}\mathrm{integer}\mathrm{finction}$$

Answered by issac last updated on 07/Nov/24

$$\underset{h=1}{\sum ^M}{2}^{-h}\le 1$$$$\underset{M\to \mathrm{\infty }}{\lim }\underset{h=1}{\sum ^M}{2}^{-h}\le 1\mathrm{but}\underset{h=1}{\sum ^{\mathrm{\infty }}}{2}^{-h}=1$$$$\mathrm{cus}0.99999......=1$$$$\lfloor \underset{M\to \mathrm{\infty }}{\lim }\underset{h=1}{\sum ^M}{2}^{-h}\rfloor =1$$

Commented by mr W last updated on 07/Nov/24

$$thequestionis$$$$\underset{n\to \mathrm{\infty }}{\lim }\lfloor \times \rfloor =?$$$$not\lfloor \underset{n\to \mathrm{\infty }}{\lim }\times \rfloor =?$$

Answered by lepuissantcedricjunior last updated on 09/Nov/24

$$\underset{\mathit{n}\to +\mathrm{\infty }}{\lim }\left[\underset{\mathit{r}=1}{\sum ^{\mathit{n}}}\frac{1}{2^{\mathit{r}}}\right]=\underset{\mathit{n}\to +\mathrm{\infty }}{\lim }[-1+\underset{\mathit{r}=0}{\sum ^{\mathit{n}}}\frac{1}{2^{\mathit{r}}}]$$$$=\underset{\mathit{n}\to +\mathrm{\infty }}{\lim }[-1+\frac{1-\left(\frac{1}{2^{\mathit{n}+1}}\right)}{1-\frac{1}{2}}]$$$$=\underset{\mathit{n}\to +\mathrm{\infty }}{\lim }[-1+2(1-2^{-(\mathit{n}+1)})]$$$$=[-1+2]=1$$

Commented by mr W last updated on 09/Nov/24

$$wrong!$$

Answered by mr W last updated on 07/Nov/24

$$\underset{r=1}{\sum ^n}\frac{1}{2^r}=\frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2}}=1-\frac{1}{2^n}<1,but>0$$$$\Rightarrow \left[\underset{r=1}{\sum ^n}\frac{1}{2^r}\right]=0$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }\left[\underset{r=1}{\sum ^n}\frac{1}{2^r}\right]=\underset{n\to \mathrm{\infty }}{\lim 0}=0✓$$

Commented by issac last updated on 07/Nov/24

$$Damniwaswrong$$

Commented by universe last updated on 07/Nov/24

$$thankssir$$

Answered by Berbere last updated on 07/Nov/24

$$\left[x\right]<x$$$$\Rightarrow 0⩽\left[\underset{r=1}{\sum ^n}\frac{1}{2^r}\right]⩽\underset{r=1}{\sum ^n}\frac{1}{2^r}=1-{\left(\frac{1}{2}\right)}^n<1$$$$\Rightarrow \mathrm{\forall }n\in \mathbb{N}\left[\mathrm{\Sigma }\frac{1}{2^r}\right]=0$$

Commented by universe last updated on 07/Nov/24

$$thankyousomuchsir$$

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