- Thread starter
- #1

- Thread starter dwsmith
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 890

No, you sum the infinite geometric series, and add \(33/100\). Or sum:I am given the number $.334444\ldots$

So we have $0+33\times 10^{-2} + \sum\limits_{n=3}^{\infty}4\times 10^{-n}$

Is there a way to put this all together in one geometric series?

\(\sum\limits_{n=1}^{\infty}4\times 10^{-n}=(4/10)(10/9)\)

and subtract \(11/100\)

CB

- Feb 13, 2012

- 1,704

$\displaystyle .33444...= \frac{1}{3} + \frac{1}{900} = \frac{301}{900}= \frac{1}{100}\ \frac{301}{9}= \frac{301}{1000}\ \sum_{n=0}^{\infty} 10^{-n}$I am given the number $.334444\ldots$

So we have $0+33\times 10^{-2} + \sum\limits_{n=3}^{\infty}4\times 10^{-n}$

Is there a way to put this all together in one geometric series?

Kind regards

$\chi$ $\sigma$