How do you find the sum of the infinite geometric series 4 + 3 + 2.25 + 1.6875 +… ?

1 Answer
George C.
Nov 10, 2015

Use the formula #sum_(n=1)^oo a r^(n-1) = a/(1-r)# to find #4+3+2.25+...=16#

Explanation:

The sum of a geometric series with initial term #a# and common ratio #r# is #a/(1-r)#.

To see this note that:

#(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)#

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

#=a(1-r^N)#

So if #abs(r) < 1# we have:

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

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

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

In our case #a=4# and #r=3/4# so

#sum_(n=1)^oo a r^(n-1)= a/(1-r) =4/(1-3/4)=16#

Related questions
See all questions in Infinite Series
Impact of this question
3419 views around the world
You can reuse this answer
Creative Commons License