Question

How do you find the sum of the infinite geometric series 1/2+1/4+1/8+1/16..?

Answer 1

`sum_(n=1)^oo(1/2)*(1/2)^(n-1)=1`

Explanation:

Let the series be represented by `sum_(n=1)^oo(x_n)`
The common ratio of this infinite geometric series is
`r=x_(n+1)/x_n=1/4/1/2=1/2`

Since `|r|=1/2<1`, it implies that this series converges to the value `a/(1-r)`, where `a` is the first term in the series.

`therefore sum_(n=1)^oo(1/2)*(1/2)^(n-1)=a/(1-r)=(1/2)/(1-1/2)=1`.