Investigate the value of r^n for |r|>1 and |r|<1 as n increases.
See solution.
Practice makes perfect
We already know the rule for the sum of the first n terms of a geometric series with first term a_1 and common ratio r≠ 1.
S_n=a_1 (1-r^n/1-r )
We want to write a rule for the sum of an infinite geometric series. To do so, we will show how the value of r^n changes as n increases. We will examine it in two cases.
Case I: |r| < 1
Case II: |r| > 1
Let's do it!
Case I
Consider the function f(n)=r^n, where |r|<1. This function represents an exponential decay function. Therefore, it has an horizontal asymptote at the x-axis, or y=0.
In this case where the absolute value of r is less than 1, this means r^n approaches 0 as n increases.
r^n n→∞ ⟶ 0
Since r^n approaches 0 as n increases, we can ignore r^n in the rule for S_n for large of values of n.
S_n =a_1 (1- r^n/1-r ) n→∞ ⟶ & a_1 (1- 0/1-r )
[1.2em]
&= a_1/1-r
Therefore, if the absolute value of the common ratio of an infinite geometric series is less than 1, its sum S can be calculated as follows.
S = a_1/1-r
Case II
Be aware that when the absolute value of r is greater than 1, r^n grows rapidly. With this in mind, let's consider the function f(n)=r^n, where |r|>1.
As we can see, the value of r^n becomes extremely large as n increases. Therefore, an infinite geometric series with |r|>1 does not have a finite sum.
