# Geometric series

• Jul 31st 2008, 09:41 AM
vodka
Geometric series
Express the decimal as an rational number through geometric series

1.213213213....

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

I understand how to do say .2222222222

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

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.
• Jul 31st 2008, 09:56 AM
Plato
Here is a hint to get you started.
$\displaystyle 0.213213\overline {213} = \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}}$
• Jul 31st 2008, 11:03 AM
vodka
Quote:

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

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

$\displaystyle \frac{ 120.108108} {99}\ = 1.213213213$
• Jul 31st 2008, 11:11 AM
Plato
$\displaystyle 1.\overline {213} = 1 + \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}} = 1 + \frac{{\frac{{213}}{{1000}}}}{{1 - \frac{1}{{1000}}}} = 1 + \frac{{213}}{{999}}$
• Jul 31st 2008, 11:17 AM
vodka
Quote:

Originally Posted by Plato
$\displaystyle 1.\overline {213} = 1 + \sum\limits_{n = 1}^\infty {\frac{{213}}{{10^{3n} }}} = 1 + \frac{{\frac{{213}}{{1000}}}}{{1 - \frac{1}{{1000}}}} = 1 + \frac{{213}}{{999}}$

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.