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Question Number 193866 by liuxinnan last updated on 21/Jun/23

$$prove$$$$\underset{i=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{n^i}=\frac{1}{n-1}n\in {\mathbb{N}}^{\ast }$$$$andifn>0\wedge n\in \mathbb{R}$$$$isitright?$$

Answered by deleteduser1 last updated on 21/Jun/23

$$=\frac{a}{1-r}=\frac{\frac{1}{n}}{1-\frac{1}{n}}=\frac{1}{n-1}(forn>1)$$$$For0<n<1,\frac{a}{1-r}=\frac{1}{1-n}\Rightarrow Assertionistrueonly$$$$forn>1$$

Commented by Tinku Tara last updated on 22/Jun/23

$$\mathrm{Relationship}\mathrm{holds}\mathrm{true}\mathrm{for}$$$$\mid \mathrm{r}\mid =\mid \frac{1}{n}\mid <1\Rightarrow n<-1\mathrm{or}n>1$$

Commented by liuxinnan last updated on 22/Jun/23

$$Ithinkthatisturewhenn=1$$$$because\underset{i=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{1^i}=\underset{i=1}{\sum ^{+\mathrm{\infty }}}1=+\mathrm{\infty }$$$$\underset{n\to 1}{\lim }\frac{1}{n-1}=+\mathrm{\infty }$$

Commented by Tinku Tara last updated on 22/Jun/23

$$\mathrm{Limit}\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{for}\frac{1}{\mathrm{n}-1}atn=1$$$$\underset{n\to 1^{-}}{\lim }\frac{1}{n-1}=-\mathrm{\infty }$$$$\underset{n\to 1^{+}}{\lim }\frac{1}{n-1}=+\mathrm{\infty }$$$$\mathrm{For}\mathrm{formula}\mathrm{to}\mathrm{be}\mathrm{considered}\mathrm{valid}$$$$\mathrm{equality}\mathrm{should}\mathrm{also}\mathrm{hold}.$$$$\mathrm{Equality}\mathrm{does}\mathrm{not}\mathrm{for}\mathrm{n}=1.$$$$\frac{1}{n-1}\ne \mathrm{\infty }$$

Commented by liuxinnan last updated on 22/Jun/23

$$yesyouareright$$

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