# 3. Infinite Geometric Series

by M. Bourne

If `-1 < r < 1`, then the infinite geometric series

a_{1}+a_{1}r+a_{1}r^{2}+a_{1}r^{3}+ ... +a_{1}r^{n-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 *a*_{1} = 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:

a_{1}= 5

a_{1}+a_{1}r= 5 + 2.5 = 7.5

a_{1}+a_{1}r+a_{1}r^{2}= 5 + 2.5 + 1.25 = 8.75

a_{1}+a_{1}r+a_{1}r^{2}+a_{1}r^{3}= 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`

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