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

Oct 22, 2017

$\textcolor{b l u e}{\frac{1}{9}}$

#### Explanation:

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

Find the common ratio, by calculating the first three terms.

${\left(\frac{1}{10}\right)}^{1} , {\left(\frac{1}{10}\right)}^{2} , {\left(\frac{1}{10}\right)}^{3} = \frac{1}{10} , \frac{1}{100} , \frac{1}{1000}$

Ratio:

$\frac{\frac{1}{100}}{\frac{1}{10}} = \frac{\frac{1}{1000}}{\frac{1}{100}} = \frac{1}{10}$

This could have been seen from the summation expression.

The sum of a geometric series is:

$a \left(\frac{1 - {r}^{n}}{1 - r}\right)$

Where $a$ is the first term, $r$ is the common ratio and $n$ is the nth term.

So:

$\frac{1}{10} \left(\frac{1 - 0}{1 - \left(\frac{1}{10}\right)}\right) = \frac{1}{10} \left(\frac{1}{\frac{1}{\frac{9}{10}}}\right) = \frac{1}{10} \left(\frac{10}{9}\right) = \textcolor{b l u e}{\frac{1}{9}}$

This could also have been arrived at using the limit to $\infty$

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