I have seen this from a book: [MATH]\sum_{n=1}^N\frac{1}{r^n}\leqslant\frac{r}{r-\delta}[/MATH] , where [MATH]\delta\lt[/MATH] r.
Now the RHS is equivalent to : [MATH]\frac{r}{r}\sum_{n=0}^\infty\left(\frac{\delta}{r}\right)^n[/MATH], so can I replace the [MATH]r[/MATH] in the LHS with [MATH]\delta\over r[/MATH]?
Since : [MATH]\frac{r}{r\left(1-{\delta\over r }\right)}=\sum_{n=0}^\infty\left(\frac{\delta}{r}\right)^n[/MATH]. We don't know whether either r or delta are greater or less than 1.
Now the RHS is equivalent to : [MATH]\frac{r}{r}\sum_{n=0}^\infty\left(\frac{\delta}{r}\right)^n[/MATH], so can I replace the [MATH]r[/MATH] in the LHS with [MATH]\delta\over r[/MATH]?
Since : [MATH]\frac{r}{r\left(1-{\delta\over r }\right)}=\sum_{n=0}^\infty\left(\frac{\delta}{r}\right)^n[/MATH]. We don't know whether either r or delta are greater or less than 1.