Question

How do you find the sum of the infinite geometric series `(2/9)^2 - (2/9)^3 + (2/9)^4 - (2/9)^5 + ......`?

Answer 1

`(2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+... = 4/33`

Explanation:

Suppose:

`S = (2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+...`

Then:

`(2/9)S = (2/9)^3-(2/9)^4+(2/9)^5-(2/9)^6+...`

So we find:

`S + (2/9)S`

`= (2/9)^2-color(red)(cancel(color(black)((2/9)^3)))+color(red)(cancel(color(black)((2/9)^4)))-color(red)(cancel(color(black)((2/9)^5)))+color(red)(cancel(color(black)((2/9)^6)))-...`
`color(white)(= (2/9)^2)+color(red)(cancel(color(black)((2/9)^3)))-color(red)(cancel(color(black)((2/9)^4)))+color(red)(cancel(color(black)((2/9)^5)))-color(red)(cancel(color(black)((2/9)^6)))+...`
`=(2/9)^2`

That is:

`(11/9)S = (1+(2/9))S = S + (2/9)S = (2/9)^2 = 4/81`

Multiply both ends by `9/11` to get:

`S = 4/33`

`color(white)()`
What is missing?

There is one vital thing missing from the above calculation, namely the fact that we have assumed that:

`(2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+...` converges.

`color(white)()`
A more formal approach

The general term of a geometric series can be written:

`a_n = a*r^(n-1)`

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

Then for any positive integer `N` we find:

`(1-r) sum_(n=0)^N a_n`

`= sum_(n=0)^N a r^(n-1) - r sum_(n=0)^N a r^(n-1)`

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

`= a(1-r^N)`

If `r != 1` then we can divide both ends by `(1-r)` to get the general formula:

`sum_(n=0)^N a_n = (a(1-r^N))/(1-r)`

If `abs(r) < 1` then `lim_(N->oo) r^N = 0` and the sum converges:

`sum_(n=0)^oo a_n = lim_(N->oo) sum_(n=0)^N a_n = lim_(N->oo) (a(1-r^N))/(1-r) = a/(1-r)`

On the other hand, if `abs(r) > 1` then `lim_(N->oo) r^N` does not converge to a finite limit, so the limit of the sum as `a/(1-r)` does not hold.

In our example:

`a_n = (2/9)^2*(-2/9)^(n-1) " "` i.e. `a=(2/9)^2` and `r = -2/9`

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

`sum_(n=0)^oo a_n = ((2/9)^2)/(1-(-2/9)) = (4/81) / (11/9) = 4/33`