- #1
ozkan12
- 149
- 0
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My Questions:
1) İn both sides of inequality of (*) why we use "n", that is, why we do multiplication with "n" ?
2) in (**) by Letting $n\to\infty$ we obtain $\lim_{{n}\to{\infty}} n\left[d\left({T}^{n}x,{T}^{n+1}x\right)\right]{}^{r}=0$ How this happened ?
3) Since, $\lim_{{n}\to{\infty}} n\left[d\left({T}^{n}x,{T}^{n+1}x\right)\right]{}^{r}=0$ there exists ${n}_{1}\in\Bbb{N}$ such that $d\left({T}^{n}x, {T}^{n+1}x\right)\le\frac{1}{{n}^{\frac{1}{r}}}$... How we can write this ?
Also, why we use number $\frac{1}{{n}^{\frac{1}{r}}}$ ?
Please can you give an answer to my questions ? Thank you for your attention..
My Questions:
1) İn both sides of inequality of (*) why we use "n", that is, why we do multiplication with "n" ?
2) in (**) by Letting $n\to\infty$ we obtain $\lim_{{n}\to{\infty}} n\left[d\left({T}^{n}x,{T}^{n+1}x\right)\right]{}^{r}=0$ How this happened ?
3) Since, $\lim_{{n}\to{\infty}} n\left[d\left({T}^{n}x,{T}^{n+1}x\right)\right]{}^{r}=0$ there exists ${n}_{1}\in\Bbb{N}$ such that $d\left({T}^{n}x, {T}^{n+1}x\right)\le\frac{1}{{n}^{\frac{1}{r}}}$... How we can write this ?
Also, why we use number $\frac{1}{{n}^{\frac{1}{r}}}$ ?
Please can you give an answer to my questions ? Thank you for your attention..