0.9999...=∞(0.9)(0.1)^n-1= 0.9/1-0.1=1,

∑

n=1

using the geometric series. By the same method, compute the exact

value of the following periodic innite decimal expression as a rational number: 0:89898989

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- Nov 19th 2013, 11:28 AMDfaulk044sequences and series
0.9999...=∞(0.9)(0.1)^n-1= 0.9/1-0.1=1,

∑

n=1

using the geometric series. By the same method, compute the exact

value of the following periodic innite decimal expression as a rational number: 0:89898989

- Nov 19th 2013, 11:38 AMPlatoRe: sequences and series
That is a very very poor way to write or think about it.

First write $\displaystyle {\rm{0:89}}\overline {{\rm{89}}} = \frac{{89}}{{{{10}^2}}} + \frac{{89}}{{{{10}^4}}} + \frac{{89}}{{{{10}^6}}} + \cdots = \sum\limits_{k = 1}^\infty {\frac{{89}}{{{{10}^{2K}}}}} $

Now your first term is $\displaystyle \frac{{89}}{{{{10}^2}}}$ and the common ratio is $\displaystyle \frac{{1}}{{{{10}^2}}}$ - Nov 19th 2013, 11:43 AMebainesRe: sequences and series
First let me rewrite what I think you tried to convey:

$\displaystyle 0.999 \overline 9... = \sum _{N=1} ^\infty (0.9)(0.1)^{N-1} = \frac {0.9}{1-0.1} = 1.$

If you do long division of 1-0.1 into 0.9 you get the infinite series .9 + .01 + .009 + ....; Hence 0.99999... = 1.

For the number 0.898989.. you have:

$\displaystyle 0.8989 \overline{89}.. = \sum _{N=1} ^\infty (0.89)(0.01)^{N-1} = \frac {0.89}{1-0.01} = 89/99$.

To show this do the long division of 1 - .01 into 0.89.