# Infinite Geometric Series and Convergence

An infinite geometric series has an infinite amount of terms:

 ${a}_{1}+{a}_{1}k+{a}_{1}{k}^{2}+\cdots +{a}_{1}{k}^{n}+\cdots$

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 $\text{}\text{}5000\text{}$ to good students of mathematics, for all eternity. The money is deposited into a savings account with an annual interest rate of $\text{}3.5\text{}\phantom{\rule{0.17em}{0ex}}\text{%}$. 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

 $\frac{5000}{1.035}+\frac{5000}{1.03{5}^{2}}+\frac{5000}{1.03{5}^{3}}+\cdots \phantom{\rule{0.17em}{0ex}},$

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:

$\begin{array}{llll}\hfill S& =\frac{{a}_{1}}{1-k}=\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\frac{50\phantom{\rule{0.17em}{0ex}}000}{1.035}\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}1-\frac{1}{1.035}\phantom{\rule{0.17em}{0ex}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 142\phantom{\rule{0.17em}{0ex}}857.14\phantom{\rule{0.17em}{0ex}}\text{}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $S=\frac{{a}_{1}}{1-k}=\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\frac{50\phantom{\rule{0.17em}{0ex}}000}{1.035}\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}1-\frac{1}{1.035}\phantom{\rule{0.17em}{0ex}}}\approx 142\phantom{\rule{0.17em}{0ex}}857.14\phantom{\rule{0.17em}{0ex}}\text{}.$

Thus, the unknown multimillionaire has to deposit $142\phantom{\rule{0.17em}{0ex}}857.14$ \$ in the account.

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

 $\phantom{\rule{-0.17em}{0ex}}|k\left(x\right)|<1$

or by solving

 $k{\left(x\right)}^{2}<1.$

The two procedures are equal.

Rule

### Convergencearea

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

 $k{\left(x\right)}^{2}<1.$

Example 2

### Option1:Solvingwithtwocasesofinequality

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

You begin by setting up the inequality:

 $\phantom{\rule{-0.17em}{0ex}}|k\left(x\right)|=\phantom{\rule{-0.17em}{0ex}}|2x+3|<1$

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.

$\begin{array}{llll}\hfill 2x+3& <1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x& <-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& <-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill -\left(2x+3\right)& <1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x-3& <1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -2x& <4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& >-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llllllll}\hfill 2x+3& <1\phantom{\rule{2em}{0ex}}& \hfill -\left(2x+3\right)& <1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x& <-2\phantom{\rule{2em}{0ex}}& \hfill -2x-3& <1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& <-1\phantom{\rule{2em}{0ex}}& \hfill -2x& <4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill x& >-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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.

From the sign lines you see can that the series converges when $x\in \phantom{\rule{-0.17em}{0ex}}\left(-2,-1\right)$.

Example 3

### Option2:Solvingwithsingleinequality

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

You can also solve the exercise by looking at the inequality

 $k{\left(x\right)}^{2}<1.$

In this case you get that $\begin{array}{llll}\hfill {\left(2x+3\right)}^{2}& <1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(2x+3\right)}^{2}-1& <0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can use the third algebraic identity to factorize the left-hand side: $\begin{array}{llll}\hfill \left(\left(2x+3\right)+1\right)\left(\left(2x+3\right)-1\right)& <0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left(2x+4\right)\left(2x+2\right)& <0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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

Because you are looking for the area where $k{\left(x\right)}^{2}<1$, the answer is the interval with the dashed line. That gives you that the series converges when $x\in \phantom{\rule{-0.17em}{0ex}}\left(-2,-1\right)$.