prove-i-1-1-n-i-1-n-1-n-N-and-if-n-gt-0-n-R-is-it-right-

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

prove Σ_(i=1) ^(+∞) (1/n^i )=(1/(n−1)) n∈N^∗ and if n>0∧ n∈R is it right?

$${prove} \\ $$$$\:\:\underset{{i}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{i}} }=\frac{\mathrm{1}}{{n}−\mathrm{1}}\:\:\:\:\:\:{n}\in\mathbb{N}^{\ast} \\ $$$${and}\:{if}\:{n}>\mathrm{0}\wedge\:{n}\in\mathbb{R} \\ $$$${is}\:{it}\:{right}? \\ $$

Answered by AST last updated on 21/Jun/23

=(a/(1−r))=((1/n)/(1−(1/n)))=(1/(n−1)) (for n>1) For 0<n<1,(a/(1−r))=(1/(1−n))⇒Assertion is true only for n>1

$$=\frac{{a}}{\mathrm{1}−{r}}=\frac{\frac{\mathrm{1}}{{n}}}{\mathrm{1}−\frac{\mathrm{1}}{{n}}}=\frac{\mathrm{1}}{{n}−\mathrm{1}}\:\left({for}\:{n}>\mathrm{1}\right) \\ $$$${For}\:\mathrm{0}<{n}<\mathrm{1},\frac{{a}}{\mathrm{1}−{r}}=\frac{\mathrm{1}}{\mathrm{1}−{n}}\Rightarrow{Assertion}\:{is}\:{true}\:{only} \\ $$$${for}\:{n}>\mathrm{1} \\ $$

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

Relationship holds true for ∣r∣=∣(1/n)∣<1 ⇒n<−1 or n>1

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

Commented by liuxinnan last updated on 22/Jun/23

I think that is ture when n=1 because Σ_(i=1) ^(+∞) (1/1^i )=Σ_(i=1) ^(+∞) 1=+∞ lim_(n→1) (1/(n−1))=+∞

$${I}\:{think}\:{that}\:{is}\:{ture}\:{when}\:{n}=\mathrm{1}\: \\ $$$${because}\:\underset{{i}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{1}^{{i}} }=\underset{{i}=\mathrm{1}} {\overset{+\infty} {\sum}}\mathrm{1}=+\infty \\ $$$$\underset{{n}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{n}−\mathrm{1}}=+\infty \\ $$

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

Limit does not exist for (1/(n−1)) at n=1 lim_(n→1^− ) (1/(n−1))=−∞ lim_(n→1^+ ) (1/(n−1))=+∞ For formula to be considered valid equality should also hold. Equality does not for n=1. (1/(n−1))≠∞

$$\mathrm{Limit}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}\:\mathrm{for}\:\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}\:{at}\:{n}=\mathrm{1} \\ $$$$\underset{{n}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}−\mathrm{1}}=−\infty \\ $$$$\underset{{n}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}−\mathrm{1}}=+\infty \\ $$$$\mathrm{For}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{be}\:\mathrm{considered}\:\mathrm{valid} \\ $$$$\mathrm{equality}\:\mathrm{should}\:\mathrm{also}\:\mathrm{hold}. \\ $$$$\mathrm{Equality}\:\mathrm{does}\:\mathrm{not}\:\mathrm{for}\:\mathrm{n}=\mathrm{1}. \\ $$$$\frac{\mathrm{1}}{{n}−\mathrm{1}}\neq\infty \\ $$

Commented by liuxinnan last updated on 22/Jun/23

$${yes}\:{you}\:{are}\:{right} \\ $$

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