# How do you find the sum of the infinite series Sigma2(1/10)^k from k=1 to oo?

Feb 10, 2017

${\sum}_{n = 1}^{\infty} 2 {\left(\frac{1}{10}\right)}^{n} = \frac{2}{9}$

#### Explanation:

Start from the geometric series of ratio $r = \frac{1}{10}$:

${\sum}_{n = 0}^{\infty} {\left(\frac{1}{10}\right)}^{n} = \frac{1}{1 - \frac{1}{10}} = \frac{10}{9}$

Extract from the sum the term for $n = 0$:

$1 + {\sum}_{n = 1}^{\infty} {\left(\frac{1}{10}\right)}^{n} = \frac{10}{9}$

${\sum}_{n = 1}^{\infty} {\left(\frac{1}{10}\right)}^{n} = \frac{10}{9} - 1 = \frac{1}{9}$

Multiplying by $2$ term by term:

${\sum}_{n = 1}^{\infty} 2 {\left(\frac{1}{10}\right)}^{n} = 2 {\sum}_{n = 1}^{\infty} {\left(\frac{1}{10}\right)}^{n} = \frac{2}{9}$