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

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)#