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

##### 1 Answer

#### 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

#S = 4/33#

**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.

**A more formal approach**

The general term of a geometric series can be written:

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

where

Then for any positive integer

#(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

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

If

#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

In our example:

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

So

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