# How do you find the sum of the infinite geometric series 0.5 + 0.05 + 0.005 + ...?

We notice that

0.5 + 0.05 + 0.005 + ...=5*[1/10+1/100+1/1000+...]= 5*[1/10+1/10^2+1/10^3+...]

But $\left[\frac{1}{10} + \frac{1}{10} ^ 2 + \frac{1}{10} ^ 3 + \ldots\right]$ it is a geometric progression with
first term ${a}_{1} = \frac{1}{10}$ and ratio $r = \frac{1}{10}$

hence its sum is given by

$S = {a}_{1} \cdot \frac{1}{1 - r} = \frac{1}{10} \cdot \frac{1}{1 - \frac{1}{10}} = \frac{1}{9}$

So the initial sum is $\frac{5}{9}$

Feb 21, 2016

S_∞ = 5/9

#### Explanation:

The sum to n terms of a geometric sequence is

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

As  n → ∞ " then " r^n → 0

and  S_∞ = a/(1 - r ) [ for -1 < r < 1 ]

where a , is the first term and r , the common ratio

here a = $0.5 = \frac{1}{2} \text{ and } r = \frac{0.5}{0.5} = \frac{0.005}{0.05} = \frac{1}{10}$

rArr S_∞ =( 1/2)/(1 - 1/10) =( 1/2)/(9/10) = 1/2xx10/9 = 5/9