Question

How do you find the sum of the infinite geometric series 4 + 3 + 2.25 + 1.6875 +… ?

Answer 1

Use the formula `sum_(n=1)^oo a r^(n-1) = a/(1-r)` to find `4+3+2.25+...=16`

Explanation:

The sum of a geometric series with initial term `a` and common ratio `r` is `a/(1-r)`.

To see this note that:

`(1-r) sum_(n=1)^N a r^(n-1) = sum_(n=1)^N a r^(n-1) - r sum_(n=1)^N a r^(n-1)`

`=sum_(n=1)^N a r^(n-1) - sum_(n=2)^(N+1) a r^(n-1)`

`=a + color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - a r^N`

`=a(1-r^N)`

So if `abs(r) < 1` we have:

`(1-r) sum_(n=1)^oo a r^(n-1)= lim_(N->oo) (1-r) sum_(n=1)^N a r^(n-1) = lim_(N->oo) a(1-r^N) = a*1 = a`

Dividing both ends by `(1-r)` we get:

`sum_(n=1)^oo a r^(n-1)=a/(1-r)`

In our case `a=4` and `r=3/4` so

`sum_(n=1)^oo a r^(n-1)= a/(1-r) =4/(1-3/4)=16`