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

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#