3. Infinite Geometric Series
by M. Bourne
If `-1 < r < 1`, then the infinite geometric series
a1 + a1r + a1r2 + a1r3 + ... + a1rn-1
converges to a particular value.
This value is given by:
`S_oo=(a_1)/(1-r)\ (|r|<1)`
The series converges because each term gets smaller and smaller (since -1 < r < 1).
Example 1
For the series:
`5 + 2.5 + 1.25 + 0.625 + 0.3125... `,
the first term is given by a1 = 5 and the common ratio is r = 0.5.
Since the common ratio has value between `-1` and `1`, we know the series will converge to some value.
Let's do the sum of the first few terms:
a1 = 5
a1 + a1r = 5 + 2.5 = 7.5
a1 + a1r + a1r2 = 5 + 2.5 + 1.25 = 8.75
a1 + a1r + a1r2 + a1r3 = 5 + 2.5 + 1.25 + 0.625 = 9.375
Continuing this pattern, we will get the following sums (correct to 9 decimal places):
Sum to 5 terms `= 9. 6875`
Sum to 6 terms `= 9. 84375`
Sum to 7 terms `= 9. 921875`
Sum to 8 terms `= 9. 9609375`
Sum to 9 terms `= 9. 98046875`
Sum to 10 terms `= 9. 990234375`
Sum to 11 terms `= 9. 995117188`
Sum to 12 terms `= 9. 997558594`
Sum to 13 terms `= 9. 998779297`
Sum to 14 terms `= 9. 999389648`
Where do we use this?
See in a later chapter how we use the sum of an infinite GP and differentiation to find polynomial approximations for functions.
We also see how a calculator works, using these progressions.
We could keep going and would see that the sum gets closer and closer to, but does not go over `10`. It appears the infinite sum is 10.
Applying the formula now, we confirm our result:
`S_oo=(a_1)/(1-r)=5/(1-0.5)=5/0.5=10`
Example 2
Find the value of the infinite geometric series:
`4+2+1+1/2+1/4+1/8+...`
Answer
Here,
`a_1=4,\ r=1/2`
`s_oo=(a_1)/(1-r)`
`S_oo=4/(1-1/2)=4/(1/2)=8`