We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra CoRequisite Course

Geometric Series

Learning Outcomes

  • Write the first n terms of a geometric sequence.
  • Determine whether the sum of an infinite geometric series exists.
  • Give the sum of a convergent infinite geometric series.
  • Solve an annuity problem using a geometric series.
Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, rr. We can write the sum of the first nn terms of a geometric series as

Sn=a1+a1r+a1r2+...+a1rn−1{S}_{n}={a}_{1}+{a}_{1}r+{a}_{1}{r}^{2}+...+{a}_{1}{r}^{n - 1}.

Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first nn terms of a geometric series. We will begin by multiplying both sides of the equation by rr.

rSn=a1r+a1r2+a1r3+...+a1rnr{S}_{n}={a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+...+{a}_{1}{r}^{n}

Next, we subtract this equation from the original equation.

\begin{align}{S}_{n}&={a}_{1}+{a}_{1}r+{a}_{1}{r}^{2}+...+{a}_{1}{r}^{n - 1} \\ -r{S}_{n}&=-\left({a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+...+{a}_{1}{r}^{n}\right) \\ \hline \left(1-r\right){S}_{n}&={a}_{1}-{a}_{1}{r}^{n}\end{align}

Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for Sn{S}_{n}, factor a1a_1 on the right hand side and divide both sides by (1−r)\left(1-r\right).

Sn=a1(1−rn)1−r r≠1{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}

tip for success

The formulas for finding the sum of the first nn terms of arithmetic and geometric series are handy and should be memorized although it is important to understand how they were derived.

A General Note: Formula for the Sum of the First n Terms of a Geometric Series

A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first nn terms of a geometric sequence is represented as

Sn=a1(1−rn)1−r r≠1{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}

How To: Given a geometric series, find the sum of the first n terms.

  1. Identify a1,r,andn{a}_{1},r,\text{and}n.
  2. Substitute values for a1,r{a}_{1},r, and nn into the formula Sn=a1(1−rn)1−r{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}.
  3. Simplify to find Sn{S}_{n}.

Example: Finding the First n Terms of a Geometric Series

Use the formula to find the indicated partial sum of each geometric series.
  1. S11{S}_{11} for the series 8+−4+2+… 8 + -4 + 2 + \dots
  2. ∑k=163⋅2k\sum\limits _{k=1}^6 3\cdot {2}^{k}

Answer:

  1. a1=8{a}_{1}=8, and we are given that n=11n=11.We can find rr by dividing the second term of the series by the first. r=−48=−12r=\dfrac{-4}{8}=-\frac{1}{2} Substitute values for a1,r,andn{a}_{1}, r, \text{and} n into the formula and simplify. \begin{align}&{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r} \\[1mm] &{S}_{11}=\dfrac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336 \\ \text{ } \end{align}
  2. Find a1{a}_{1} by substituting k=1k=1 into the given explicit formula. a1=3⋅21=6{a}_{1}=3\cdot {2}^{1}=6 We can see from the given explicit formula that r=2r=2. The upper limit of summation is 6, so n=6n=6.Substitute values for a1,r{a}_{1},r, and nn into the formula, and simplify. \begin{align}\\ &{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r} \\[1mm] &{S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1 - 2}=378 \end{align}

Try It

Use the formula to find the indicated partial sum of each geometric series. S20[/latex]fortheseries[latex]1,000+500+250+…{S}_{20}[/latex] for the series [latex]1\text{,}000 + 500 + 250 + \dots

Answer: ≈2,000.00\approx 2,000.00

  Use the formula to determine the sum ∑k=183k\sum\limits _{k=1}^{8}{3}^{k}

Answer: 9,840

[embed]

Example: Solving an Application Problem with a Geometric Series

At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.

Answer: The problem can be represented by a geometric series with a1=26,750{a}_{1}=26,750; n=5n=5; and r=1.016r=1.016. Substitute values for a1{a}_{1}, rr, and nn into the formula and simplify to find the total amount earned at the end of 5 years.

\begin{align}{S}_{n}&=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r} \\ {S}_{5}&=\dfrac{26\text{,}750\left(1-{1.016}^{5}\right)}{1 - 1.016}\approx 138\text{,}099.03 \end{align}

He will have earned a total of $138,099.03 by the end of 5 years.

Try It

At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?

Answer: $275,513.31

[embed]

Using the Formula for the Sum of an Infinite Geometric Series

Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first n terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is 2+4+6+8+…2+4+6+8+\dots. This series can also be written in summation notation as ∑k=1∞2k \sum\limits _{k=1}^{\infty} 2k, where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges.

Determining Whether the Sum of an Infinite Geometric Series is Defined

If the terms of an infinite geometric series approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:

1+0.2+0.04+0.008+0.0016+…1+0.2+0.04+0.008+0.0016+\dots

The common ratio is r=0.2r=0.2. As n gets large, the values of of rnr^n get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with −1<r<1-1<r<1 approach 0; the sum of a geometric series is defined when −1<r<1-1<r<1.

DETERMINING WHETHER THE SUM OF AN INFINITE GEOMETRIC SERIES IS DEFINED

The sum of an infinite series is defined if the series is geometric and −1<r<1-1<r<1.

How To: Given the first several terms of an infinite series, determine if the sum of the series exists.

  1. Find the ratio of the second term to the first term.
  2. Find the ratio of the third term to the second term.
  3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.
  4. If a common ratio, r, was found in step 3, check to see if −1<r<1-1<r<1. If so, the sum is defined. If not, the sum is not defined.

tip for success

Remember to try the examples below before looking at the answers!

Example: Determining Whether the Sum of an Infinite Series is Defined

Determine whether the sum of each infinite series is defined.
  1. 12+8+4+…12+8+4+\dots
  2. 34+12+13+…\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{1}{3}+\dots
  3. ∑k=1∞27⋅(13)k\sum\limits _{k=1}^{\infty}{27}\cdot\left(\dfrac{1}{3}\right)^k
  4. ∑k=1∞5k\sum\limits _{k=1}^{\infty}{5k}

Answer:

  1. The ratio of the second term to the first is 23\frac{2}{3}, which is not the same as the ratio of the third term to the second, 12\frac{1}{2}. The series is not geometric.
  2. The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of 23\frac{2}{3}. The sum of the infinite series is defined.
  3. The given formula is exponential with a base of 13\frac{1}{3}; the series is geometric with a common ratio of 13\frac{1}{3}. The sum of the infinite series is defined.
  4. The given formula is not exponential. The series is arithmetic, not geometric and so cannot yield a finite sum.

try it

Determine whether the sum of the infinite series is defined.
  1. 13+12+34+98+⋯\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{9}{8}+\cdots
  2. 24+(−12)+6+(−3)+…24+(-12)+6+(-3)+\dots
  3. ∑k=1∞15⋅(−0.3)k\sum\limits _{k=1}^{\infty} 15\cdot(-0.3)^k

Answer:

  1. The series is geometric, but r=32>1r=\dfrac{3}{2}>1. The sum is not defined.
  2. The series is geometric with r=−12r=-\dfrac{1}{2}. The sum is defined.
  3. The series is geometric with r=−0.3r=-0.3. The sum is defined.

Finding Sums of Infinite Series

When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first n terms of a geometric series.

Sn=a1(1−rn)1−r{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}

We will examine an infinite series with r=12r=\frac{1}{2}. What happens to rnr^n as n increases?

\begin{align} &{\left(\frac{1}{2}\right)}^{2} = \frac{1}{4} \\&{\left(\frac{1}{2}\right)}^{3} = \frac{1}{8} \\&{\left(\frac{1}{2}\right)}^{4} = \frac{1}{16} \end{align}

The value of rnr^n decreases rapidly. What happens for greater values of n?

\begin{align} &{\left(\frac{1}{2}\right)}^{10} = \frac{1}{1\text{,}024} \\&{\left(\frac{1}{2}\right)}^{20} = \frac{1}{1\text{,}048\text{,}576} \\&{\left(\frac{1}{2}\right)}^{30} = \frac{1}{1\text{,}073\text{,}741\text{,}824} \end{align}

As n gets large, rnr^n gets very small. We say that as n increases without bound, rnr^n approaches 0. As rnr^n approaches 0, 1−rn1-r^n approaches 1. When this happens the numerator approaches a1a_1. This gives us the formula for the sum of an infinite geometric series.

tip for success

This is another very handy formula that should be memorized. It is important to understand, though, how it is derived from the formula for the sum of the first nn terms of a geometric series for nn increasing without bound.

A General Note: FORMULA FOR THE SUM OF AN INFINITE GEOMETRIC SERIES

The formula for the sum of an infinite geometric series with −1<r<1-1<r<1 is:

S=a11−rS=\dfrac{{a}_{1}}{1-r}

How To: Given an infinite geometric series, find its sum.

  1. Identify a1a_1 and r.
  2. Confirm that −1<r<1-1<r<1.
  3. Substitute values for a1a_1 and r into the formula, S=a11−rS=\dfrac{{a}_{1}}{1-r}.
  4. Simplify to find S.

Example: Finding the Sum of an Infinite Geometric Series

Find the sum, if it exists, for the following:
  1. 10+9+8+7+…10+9+8+7+\dots
  2. 248.6+99.44+39.776+…248.6+99.44+39.776+\dots
  3. ∑k=1∞4,374⋅(−13)k−1\sum\limits _{k=1}^{\infty}4\text{,}374\cdot\left(-\dfrac{1}{3}\right)^{k-1}
  4. ∑k=1∞19⋅(43)k\sum\limits _{k=1}^{\infty}\dfrac{1}{9}\cdot\left(\dfrac{4}{3}\right)^{k}

Answer:

  1. There is not a constant ratio; the series is not geometric.
  2. There is a constant ratio; the series is geometric. a1=248.6a_1=248.6 and r=99.44248.6=0.4r=\dfrac{99.44}{248.6}=0.4, so the sum exists. Substitute a1=248.6a_1=248.6 and r=0.4r=0.4 into the formula and simplify to find the sum. \begin{align} \\ &S=\frac{a_1}{1-r} \\[1.5mm] &S=\frac{248.6}{1-0.4}=\frac{1243}{3} \\ \text{ }\end{align}
  3. The formula is exponential, so the series is geometric with r=−13r=-\frac{1}{3}. Find a1a_1 by substituting k=1k=1 into the given explicit formula. a1=4,374⋅(−13)1−1=4,374 \begin{align} \\ a_1=4\text{,}374\cdot\left(-\frac{1}{3}\right)^{1-1}=4\text{,}374 \\ \text{ }\end{align} Substitute 4,3744\text{,}374 and r=−13r=-\frac{1}{3} into the formula, and simplify to find the sum. \begin{align}\\&S=\frac{a_1}{1-r} \\[1.5mm] &S=\frac{4\text{,}374}{1-\left(-\frac{1}{3}\right)}=3\text{,}280.5 \\ \text{ }\end{align}
  4. The formula is exponential, so the series is geometric, but r>1r>1. The sum does not exist.

Example: Finding an Equivalent Fraction for a Repeating Decimal

Find the equivalent fraction for the repeating decimal 0.3‾0.\overline{3}.

Answer: We notice the repeating decimal 0.3‾=0.333…0.\overline{3}=0.333\dots. so we can rewrite the repeating decimal as a sum of terms.

0.3‾=0.3+0.03+0.003+…0.\overline{3}=0.3+0.03+0.003+\dots

Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term.

\begin{align}0.\overline{3}&=0.3+0.3\cdot(0.1)+0.3\cdot(0.01)+0.3\cdot(0.001)+\dots \\ &=0.3+0.3\cdot(0.1)+0.3\cdot(0.1)^2+0.3\cdot(0.1)^3+\dots\end{align}

Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have

S=a11−r=0.31−0.1=0.30.9=13S=\dfrac{a_1}{1-r} =\dfrac{0.3}{1-0.1} =\dfrac{0.3}{0.9} =\dfrac{1}{3}

try it

Find the sum if it exists.
  1. 2+23+29+…2+\dfrac{2}{3}+\dfrac{2}{9}+\dots
  2. ∑k=1∞0.76k+1\sum\limits _{k=1}^{\infty}{0.76k+1}
  3. ∑k=1∞(−38)k\sum\limits _{k=1}^{\infty}\left(-\dfrac{3}{8}\right)^k

Answer:

  1.  3
  2.  The series is arithmetic. The sum does not exist.
  3.  −311-\dfrac{3}{11}

[ohm_question hide_question_numbers=1]20285[/ohm_question] [ohm_question hide_question_numbers=1]20287[/ohm_question]

Annuities

At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added. We can find the value of the annuity right after the last deposit by using a geometric series with a1=50{a}_{1}=50 and r=100.5%=1.005r=100.5\%=1.005. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned. We can find the value of the annuity after nn deposits using the formula for the sum of the first nn terms of a geometric series. In 6 years, there are 72 months, so n=72n=72. We can substitute a1=50,r=1.005,{a}_{1}=50, r=1.005, and n=72n=72 into the formula, and simplify to find the value of the annuity after 6 years.

S72=50(1−1.00572)1−1.005≈4,320.44{S}_{72}=\dfrac{50\left(1-{1.005}^{72}\right)}{1 - 1.005}\approx 4\text{,}320.44

After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of 72\left(50\right) = $3,600. This means that because of the annuity, the couple earned $720.44 interest in their college fund.

How To: Given an initial deposit and an interest rate, find the value of an annuity.

  1. Determine a1{a}_{1}, the value of the initial deposit.
  2. Determine nn, the number of deposits.
  3. Determine rr.
    1. Divide the annual interest rate by the number of times per year that interest is compounded.
    2. Add 1 to this amount to find rr.
  4. Substitute values for a1,r,{a}_{1},r, and nn into the formula for the sum of the first nn terms of a geometric series, Sn=a1(1−rn)1−r{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}.
  5. Simplify to find Sn{S}_{n}, the value of the annuity after nn deposits.

Example: Solving an Annuity Problem

A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?

Answer: The value of the initial deposit is $100, so a1=100{a}_{1}=100. A total of 120 monthly deposits are made in the 10 years, so n=120n=120. To find rr, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.

r=1+0.0912=1.0075r=1+\dfrac{0.09}{12}=1.0075

Substitute a1=100,r=1.0075,{a}_{1}=100,r=1.0075, and n=120n=120 into the formula for the sum of the first nn terms of a geometric series, and simplify to find the value of the annuity.

S120=100(1−1.0075120)1−1.0075≈19,351.43{S}_{120}=\dfrac{100\left(1-{1.0075}^{120}\right)}{1 - 1.0075}\approx 19\text{,}351.43

So the account has $19,351.43 after the last deposit is made.

Try It

At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?

Answer: $92,408.18

[embed]
 

Licenses & Attributions