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Infinite Geometric Series and Convergence

An infinite geometric series has an infinite amount of terms:

a1 + a1k + a1k2 + + a 1kn +

The sum of the series converges towards a particular number if the quotient k is between 1 and 1. In that case, the sum is

S = a1 1 k.

Example 1

An endowment is generated by a retired multimillionaire which is giving out an annual scholarship of $5000 to good students of mathematics, for all eternity. The money is deposited into a savings account with an annual interest rate of 3.5%. What amount of money needs to be deposited into this account?

The present values of the annual payments to the mathematics student forms the infinite geometric series

5000 1.035 + 5000 1.0352 + 5000 1.0353 + ,

where k = 1 1.035 and a1 = 5000 1.035. You know that the series is converging because k is between 1 and 1.

You have to find the sum of the an infinite geometric series to figure out how much that was deposited into the bank account to begin with:

S = a1 1 k = 50000 1.035 1 1 1.035 142857.14$.

S = a1 1 k = 50000 1.035 1 1 1.035 142857.14$.

Thus, the unknown multimillionaire has to deposit 142857.14 $ in the account.

When the quotient k is a function of x, the convergence area is given by 1 < k(x) < 1. Then you can find the convergence area either by solving

|k(x)| < 1

or by solving

k(x)2 < 1.

The two procedures are equal.

Rule

Convergence area

When the quotient k is a function of x, the convergence area is given by 1 < k(x) < 1. Then you can find the convergence area by solving

k(x)2 < 1.

Example 2

Option 1: Solving with two cases of inequality

You have a geometric series with the quotient k(x) = 2x + 3. Find the area where it converges.

You begin by setting up the inequality:

|k(x)| = |2x + 3| < 1

Since this is an absolute value, you have to divide it into two inequalities. Solve them separately and use a sign chart to find the interval you are looking for.

2x + 3 < 1 2x < 2 x < 1 (2x + 3) < 1 2x 3 < 1 2x < 4 x > 2

2x + 3 < 1 (2x + 3) < 1 2x < 2 2x 3 < 1 x < 1 2x < 4 x > 2

Draw the sign lines for the inequalities. You find the solution by finding the area where both inequalities are true. That area is where the infinite geometric series converge.

A sign chart for the convergence area

From the sign lines you see can that the series converges when x (2,1).

Example 3

Option 2: Solving with single inequality

You have a geometric series with the quotient k(x) = 2x + 3.. Find the area where it converges.

You can also solve the exercise by looking at the inequality

k(x)2 < 1.

In this case you get that (2x + 3)2 < 1 (2x + 3)2 1 < 0

You can use the third algebraic identity to factorize the left-hand side: ((2x + 3) + 1)((2x + 3) 1) < 0 (2x + 4)(2x + 2) < 0

Now, you can use a sign chart to find the answer. Draw them and interpret the lines:

A sign chart for the convergence area

Because you are looking for the area where k(x)2 < 1, the answer is the interval with the dashed line. That gives you that the series converges when x (2,1).

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