Chapter Contents

• Series and the Binomial Theorem
• 1. Arithmetic Progressions
• 2. Geometric Progressions
• 3. Infinite Geometric Series
• 4. The Binomial Theorem
• Series and the Binomial Theorem Problem Solver

• Comments, Questions?

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3. Infinite Geometric Series

by M. Bourne

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

a₁ + a₁r + a₁r² + a₁r³ + ... + a₁r^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 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`

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