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

Continues below

Example 2

Find the value of the infinite geometric series:

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

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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