Question

How do you find the sum of the infinite geometric series given `1+2/3+4/9+...`?

Answer 1

`1+2/3+4/9+... = 3`

Explanation:

The general term of any geometric series can be written in the form:

`a_n = a*r^(n-1)" "` for `n = 1, 2, 3,...`

where `a` is the initial term and `r` the common ratio

In our case we have:

`a_n = 1*(2/3)^(n-1)" "` for `n = 1, 2, 3,...`

with initial term `a=1` and common ratio `r=2/3`

The general formula for the infinite sum (proved below) is:

`sum_(n=1)^oo ar^(n-1) = a/(1-r)` when `abs(r) < 1`

So in our case:

`sum_(n=1)^oo color(blue)(1)*(color(purple)(2/3))^(n-1) = color(blue)(1)/(1-color(purple)(2/3)) = 1/(1/3) = 3`

`color(white)()`
Background

The general term of a geometric series can be written:

`a_n = a*r^(n-1)" "` for `n = 1, 2, 3,...`

where `a` is the initial term and `r` is the common ratio.

Given such a series, we find:

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

`color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)`

`color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)`

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

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

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

Dividing both ends by `(1-r)` we get the general finite sum formula:

`color(blue)(sum_(n=1)^N ar^(n-1) = (a(1-r^N))/(1-r))`

If `abs(r) < 1` then `lim_(N->oo) r^N = 0` and we find:

`sum_(n=1)^oo ar^(n-1) = lim_(N->oo) sum_(n=1)^N ar^(n-1)`

`color(white)(sum_(n=1)^oo ar^(n-1)) = lim_(N->oo) (a(1-r^N))/(1-r)`

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

So we have the general formula for the infinite sum:

`color(blue)(sum_(n=1)^oo ar^(n-1) = a/(1-r))" "` when `color(blue)(abs(r) < 1)`