How do you calculate #sum_(n=1)^(∞)(3/2)^(1-2n)#?

Note that:

#sum_(n=1)^oo (3/2)^(1-2n) = sum_(n=1)^oo (2/3) (3/2)^(2-2n)#

#color(white)(sum_(n=1)^oo (3/2)^(1-2n)) = sum_(n=1)^oo (2/3) (2/3)^(2n-2)#

#color(white)(sum_(n=1)^oo (3/2)^(1-2n)) = sum_(n=1)^oo (2/3) (4/9)^(n-1)#

This is now in the standard form of a geometric series with initial term #a=2/3# and common ratio #4/9#.

Given any geometric series, the general term can be written:

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

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

Then we find:

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

#color(white)((1-r) sum_(n=1)^N a r^(n-1)) = sum_(n=1)^N a r^(n-1) - sum_(n=2)^(N+1) a r^(n-1)#

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

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

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

Dividing both ends by #(1-r)# we find:

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

So if #abs(r) < 1# then:

#sum_(n=1)^oo a r^(n-1) = lim_(N->oo) sum_(n-1)^N a r^(n-1) = lim_(N->oo) (a(1-r^N))/(1-r) = a/(1-r)#

Applying this in our example with #a=2/3# and #r=4/9# we find:

#sum_(n=1)^oo (3/2)^(1-2n) = sum_(n=1)^oo (2/3) (4/9)^(n-1) = (2/3)/(1-4/9) = (2/3)/(5/9) = 6/5#