# Find the sum of the infinite geometric series 8 + 4 + 2 + 1 +...if it exists

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Before taking the sum, determine the common ratio between consecutive terms.

`r=4/8=1/2`

`r=2/4=1/2`

Hence, the common ratio is `1/2` .

Now that the common ratio is known, plug-in its value and the first term to the formula:

`S_(oo)=a_1/(1-r)`

`S_(oo)=8/(1-1/2)=8/(1/2)`

`S_(oo)=16`

**Hence, the sum of the given infinite geometric series is 16. **

The sum of n terms of a geometric series with terms `T_n = a*r^(n-1)` is given by `S_n = a*((1- r^n)/(1-r))` . If r is less than 1, as n approaches infinity the term `r^n` approaches 0. This gives the sum of infinite terms as `a/(1- r)`

For the series 8 + 4 + 2 + 1 + ..., a = 8 and r = (1/2)

The sum of infinite terms of this series is `8/(1 - 1/2) = 8/(1/2)` = 8*2 = 16