# Math Help - Geometric series

1. ## Geometric series

Express the decimal as an rational number through geometric series

1.213213213....

$\frac{ A} {1-r} \$

I understand how to do say .2222222222

as $\frac {2} {10}\ ( 1 + \frac { 1} {10}\ + \frac { 1} {100}\ ...)$

With my A = 2/10

and my r = 1/10

Any help would be appreciated! Im sure its me just brain farting, I cant seem to put it together.

2. Here is a hint to get you started.
$0.213213\overline {213} = \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}}$

3. Originally Posted by Plato
Here is a hint to get you started.
$0.213213\overline {213} = \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}}$

Hm still lost but i did manage to fudge an answer Algerbraically though not entirely correct

by using An = 1.213213213

then multiplying both sides by 100

100An = 121.3213213

then subtracting them and eventually getting

$\frac{ 120.108108} {99}\ = 1.213213213$

4. $1.\overline {213} = 1 + \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}} = 1 + \frac{{\frac{{213}}{{1000}}}}{{1 - \frac{1}{{1000}}}} = 1 + \frac{{213}}{{999}}$

5. Originally Posted by Plato
$1.\overline {213} = 1 + \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}} = 1 + \frac{{\frac{{213}}{{1000}}}}{{1 - \frac{1}{{1000}}}} = 1 + \frac{{213}}{{999}}$
Thanks! I was not paying attention to the n=1 on the summation where on geometric series it doesn't effect the convergence/divergence just the value. Failing to realize that n=0 was my downfall. Case in point many thanks.