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. 84375

Sum to 6 terms = 9. 921875

Sum to 7 terms = 9. 9609375

Sum to 8 terms = 9. 98046875

Sum to 9 terms = 9. 990234375

Sum to 10 terms = 9. 995117188

Sum to 11 terms = 9. 997558594

Sum to 12 terms = 9. 998779297

Sum to 13 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 does not go over 10.

Applying the formula now, we get the same result:

`{:(S_oo,=(a_1)/(1-r)),(,=5/(1-0.5)),(,=5/0.5),(,=10):}`


Here is how LiveMath could add the series:

LIVEMath

Example 2

Find the value of the infinite geometric series:

`4+2+1+1/2+1/4+1/8+...`

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