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