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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From the math blog...

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. 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

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):}`

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Here is how LiveMath could add the series:

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Example 2

Find the value of the infinite geometric series:

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

Answer