Infinite Geometric Series and Convergence

An infinite geometric series has an infinite amount of terms:

a_{1} + a_{1} k + a_{1} k^{2} + \dots + a_{1} k^{n} + \dots

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 = \frac{a_{1}}{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

\frac{5000}{1.035} + \frac{5000}{1.03 5^{2}} + \frac{5000}{1.03 5^{3}} + \dots,

where $k = \frac{1}{1.035}$ and $a_{1} = \frac{5000}{1.035}$ . 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:

\begin{array}{l} S & = \frac{a_{1}}{1 - k} = \frac{\frac{50 000}{1.035}}{1 - \frac{1}{1.035}} \\ \approx 142 857.14 $ . \end{array}

S = \frac{a_{1}}{1 - k} = \frac{\frac{50 000}{1.035}}{1 - \frac{1}{1.035}} \approx 142 857.14 $ .

Thus, the unknown multimillionaire has to deposit

142 857.14

$ 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 $k (x) = 2 x + 3$ . Find the area where it converges.

You begin by setting up the inequality:

| k (x) | = | 2 x + 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.

\begin{array}{l} 2 x + 3 & < 1 \\ 2 x & < - 2 \\ x & < - 1 \\ - (2 x + 3) & < 1 \\ - 2 x - 3 & < 1 \\ - 2 x & < 4 \\ x & > - 2 \end{array}

\begin{array}{l} 2 x + 3 & < 1 & - (2 x + 3) & < 1 \\ 2 x & < - 2 & - 2 x - 3 & < 1 \\ x & < - 1 & - 2 x & < 4 \\ x & > - 2 \end{array}

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 $x \in (- 2, - 1)$ .

Example 3

Option 2: Solving with single inequality
You have a geometric series with the quotient $k (x) = 2 x + 3$ .. 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

\begin{array}{l} {(2 x + 3)}^{2} & < 1 \\ {(2 x + 3)}^{2} - 1 & < 0 \end{array}

You can use the third algebraic identity to factorize the left-hand side:

\begin{array}{l} ((2 x + 3) + 1) ((2 x + 3) - 1) & < 0 \\ (2 x + 4) (2 x + 2) & < 0 \end{array}

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 $x \in (- 2, - 1)$ .

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