A few important building blocks are necessary before working with geometric series. The first is the mathematical sequence which is simply a set of numbers with a defined order. A specific case which we will use to construct geometric series is the geometric sequence: a sequence in which each term is generated by multiplying the previous term by a fixed number called the common ratio. Any geometric sequence is defined completely by its first term and its common ratio. For instance, the sequence
$$1,2,4,8,16,\dots $$
is geometric, as each term is obtained by multiplying the previous one by a constant; in this case, the constant is 2. Note the use of the word "constant": the number by which each term is multiplied is unchanged throughout the sequence. The sequence
$$2,4,16,256\dots $$
however, is not geometric. Note that each term is the square of the one that precedes it - that is, each term is multiplied by itself, not the same number! This sequence is well defined, but it is NOT geometric.
The ability to find any term of a sequence can be a useful tool. (Here, n is indeed a variable, meaning that any particular term can be found: the 1st, 2nd, 3rd, 4th, et cetera.) The nth term will be denoted {eq}a_n {/eq}.
What is already known about this number? As each term is obtained from the previous one, it is the previous term multiplied by the common ratio, which will be denoted r. That is,
$$a_n=a_{n-1}r $$
But this number is just the previous term multiplied by r, and that previous term is just the term before it multiplied by r, and this process continues until the first term is reached. This term is usually denoted {eq}a_0 {/eq}. (Note that the subscript is zero, not one. This may seem confusing at first, but will simplify notation greatly in the future.) Therefore,
$$a_n=a_{n-1}r=a_{n-2}r^2=\cdots=a_0r^n $$
Consider the geometric sequence {eq}3,6,12,24,\dots {/eq}. What is the 10th term? Note well: the 10th term will be {eq}a_9 {/eq}, as the first term is {eq}a_0 {/eq}. With the formula above, finding it is simple once the first term and common ratio are known. The first term is simply 3; divide any term by the term before it to find the common ratio. {eq}r=6\div3=2 {/eq}. (Checking that other terms have the same ratio verifies that the sequence is indeed geometric.) The 10th term follows immediately:
$$a_9=a_0r^9=3\cdot2^9=3\cdot512=1536 $$
Writing out the sequence step by step bears out the results of the formula:
$$a_0=3\\ a_1=3\cdot2=6\\ a_2=6\cdot2=12\\ a_3=12\cdot2=24\\ a_4=24\cdot2=48\\ a_5=48\cdot2=96\\ a_6=96\cdot2=192\\ a_7=192\cdot2=384\\ a_8=384\cdot2=768\\ a_9=768\cdot2=1536 $$
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Now the question of "geometric series" can be tackled. A series is the sum of the terms of a sequence; a geometric series therefore, is the sum of the terms of a geometric sequence. For a geometric sequence {eq}a_n=a_0r^n {/eq}, the geometric series of this sequence is written
$$\sum_{i=0}^{n-1}a_0r^i $$
This "sigma notation" (the large symbol is the capital Greek letter sigma) is used as shorthand for sums. It expands to
$$a_0r^0+a_0r^1+\cdots+a_0r^{(n-1)} $$
The variable below the sigma (called the index) is replaced by 1 for the first term, 2 for the second, and so on until the number above the sigma is reached. In this case, that number is {eq}n-1 {/eq}, but if a constant is in its place, the sum is a finite sum that ends when the index reaches that number.
For instance, consider the geometric series {eq}a_n=3\cdot2^n {/eq}, which has first term 3 and common ratio 2. Its series is simply adding the terms together. In order to find the third term in the series (not the sequence), the first three terms are added together. The notation for this sum looks like this:
$$\sum_{i=0}^23\cdot2^i=3\cdot2^0+3\cdot2^1+3\cdot2^2=3\cdot1+3\cdot2+3\cdot4=3+6+12=21 $$
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Earlier in the lesson, a simpler shorthand for the {eq}n {/eq}th term of a geometric sequence was described. The same can be done for a geometric series, with a little reasoning. First, for convenience, use {eq}S_n {/eq} to denote the sum of the terms from 0 to {eq}n {/eq}:
$$S_n=\sum_{i=0}^na_0r^i $$
Proceed with a little lateral thinking: notice that each term contains {eq}r {/eq}, except for the first term. That problem can be eliminated by multiplying the entire series by {eq}r {/eq}:
$$rS_n=r\sum_{i=0}^na_0r^i=r(a_0+a_0r+\cdots+a_0r^n)=a_0r+a_0r^2+\cdots+a_0r^{(n+1)}=\sum_{i=0}^na_0r^{(i+1)} $$
And what happens when these two series are subtracted?
$$S_n-rS_n=\sum_{i=0}^na_0r^i-\sum_{i=0}^na_0r^{(i+1)}=(a_0+a_0r+\cdots+a_0r^n)-(a_0r+a_0r^2+\cdots+a_0r^{(n+1)}) $$
But wait. The second term of the first sum is the same as the first term of the second sum. Indeed, this pattern continues: the third term of the first sum is the same as the second term of the second sum, and so on. Rearrange the terms to align these identical terms together.
$$S_n-rS_n=a_0+a_0r-a_0r+a_0r^2-a_0r^2+\cdots+a_0r^n-a_0r^n-a_0r^{(n+1)} $$
These terms cancel in pairs, leaving only the last term from the first sum and the first term from the second sum.
$$S_n-rS_n=a_0-a_0r^n=a_0(1-r^{(n+1)}) $$
Solve for {eq}S_n {/eq}.
$$S_n-rS_n=a_0(1-r^{(n+1)})\\ S_n(1-r)=a_0(1-r^{(n+1)})\\ S_n=a_0\frac{(1-r^{(n+1)})}{(1-r)} $$
Now, an interesting question arises. Given any well-defined series, finding the sum of any number of its terms is a matter of calculating each term and adding them all together - provided that the number of terms is finite. What if there are infinitely many terms? Can such a sum be calculated? Does it even exist?
The answer depends upon the series. Only geometric series are considered for this lesson, for which there is a very clear answer. The formula for a sum of a finite number of terms - a partial sum - has already been defined above. Then, for an infinite series, this partial sum tends towards infinite terms.
$$\sum_{i=0}^\infty a_0r^i=\lim_{n\to\infty}S_n=\lim_{n\to\infty}a_0\frac{(1-r^{(n+1)})}{(1-r)} $$
{eq}n {/eq} only appears in one place in this limit, so that is the only term that needs consideration. Clearly, any number greater than 1 tends towards infinity as {eq}n {/eq} tends towards infinity; that is, the sum will diverge if {eq}r>1 {/eq}. However, any number less than 1 raised to a greater and greater power will grow smaller and smaller, and therefore if {eq}r<1 {/eq}, the sum converges. Note that if {eq}r=1 {/eq}, then the equation is ill defined, as the denominator becomes {eq}1-r=1-1=0 {/eq}.
The infinite sum only exists if {eq}|r|<1 {/eq}. As mentioned above, when this is the case, the term {eq}r^{(n+1)}\to0 {/eq}, so the equation becomes
$$\sum_{i=0}^\infty a_0r^i=a_0\frac{(1-0)}{(1-r)}=a_0\frac{1}{(1-r)} $$
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Below are examples of how to calculate the sum of finite and infinite geometric series.
$$\sum_{i=0}^\infty 7\cdot\frac{1}{2}^i=7\frac{1}{1-\frac{1}{2}}=7\frac{1}{\frac{1}{2}}=7\cdot2=14 $$
$$\sum_{i=0}^5 2^i=\frac{1-2^{(5+1)}}{1-2}=\frac{1-64}{1-2}=\frac{-63}{-1}=63 $$
The seventh term of the sequence
$$1,5,25\cdots $$
is calculated as such:
$$a_6=a_0r^6=1\cdot5^6=15625 $$
$$\sum_{i=0}^\infty 11\cdot\frac{4}{5}^i=11\frac{1}{1-\frac{4}{5}}=11\frac{1}{\frac{1}{5}}=11\cdot5=55 $$
$$\sum_{i=0}^3 \frac{2}{3}^i=\frac{1-\frac{2}{3}^{(3+1)}}{1-\frac{2}{3}}=\frac{1-\frac{2^4}{3^4}}{\frac{1}{3}}=3(1-\frac{16}{81})=3(\frac{65}{81})=\frac{65}{27} $$
The questions below contain both conceptual questions and numerical questions pertaining to finding the sum of a geometric series. The answers are given following the last question.
What is the difference between a sequence and a series?
Calculate the sum {eq}\sum_{i=0}^\infty (\frac{1}{2})2^i {/eq}. (Look at the formula for partial sums for a hint.)
A geometric sequence is given by {eq}1,4,16\cdots {/eq}. Find the 5th term.
Find the 4th term of the geometric series with first term 2 and common ratio 3.
When does a geometric series converge?
Find the infinite sum {eq}\sum_{i=0}^\infty \frac{3}{5}^i {/eq}.
What is the name of the constant by which each term of a geometric sequence is multiplied by? What letter usually represents it in formulae concerning geometric sequences?
$$\sum_{i=0}^4 3(\frac{1}{2})^i $$
A sequence is a set of numbers with a fixed order, while a series is a sum of the terms of a sequence.
Hint: {eq}S_n=\frac{1}{2}\frac{1-2^{(n+1)}}{1-2} {/eq}. What happens as {eq}n {/eq} approaches infinity?
This is a trick question: the sum actually diverges.
The initial value is 1, and the common ratio is clearly 4 (as {eq}1\cdot4=4 {/eq} and {eq}4\cdot4=16 {/eq}). Remember that the 5th term is denoted {eq}a_4 {/eq}, as the first term actually has subscript 0.
$$a_4=1\frac{1-4^{(4+1)}}{1-4}=\frac{1-4^5}{-3}=\frac{1-1024}{-3}=\frac{-1023}{-3}=341 $$
Note that for the 4th term, {eq}n {/eq} will be equal to 3, as the index begins at 0, not 1. From here, plug values into the formula for partial sums.
$$\sum_{i=0}^3 2\cdot3^i=2\frac{1-3^{(3+1)}}{1-3}=2\frac{1-3^4}{-2}=-\frac{2}{2}(1-80)=-1(-80)=80 $$
A geometric series converges only when {eq}|r|<1 {/eq}, where {eq}r {/eq} is the common ratio.
The formula for the sum of an infinite geometric series is
$$\frac{1}{1-r}=\frac{1}{1-\frac{3}{5}}=\frac{1}{\frac{2}{5}}=\frac{5}{2} $$
Each term of a geometric sequence is multiplied by the common ratio to find the next term. The variable used for it is {eq}r {/eq}.
Simply use the formula for partial sums.
$$\sum_{i=0}^4 3(\frac{1}{2})^i=3\frac{1-\frac{1}{2}^5}{1-\frac{1}{2}}=3\frac{1-\frac{1}{32}}{\frac{1}{2}}=(3\cdot2)\frac{31}{32}=\frac{3\cdot31}{16}=\frac{93}{16} $$
Geometric sequences are strings of numbers in which each term is produced by multiplying the previous one by a fixed common ratio {eq}r {/eq}. Geometric series are sums of these such sequences. Any individual term of a geometric sequence can be found by the formula
$$a_n=a_{n-1}r $$
and any partial sum of a geometric series by
$$S_n=a_0\frac{(1-r^{(n+1)})}{(1-r)} $$
An infinite sum is slightly more complicated. It converges only when {eq}|r|<1 {/eq}, to the formula
$$\frac{1}{1-r} $$
Besides being fascinating for creating real, finite quantities out of infinite sums, geometric series have many interesting applications. Rates of growth and decay, for instance in populations or monetary values, are often geometric.
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(a) Find the sum of the first 6 terms of the following geometric sequence:
-3, 6, -12, 24,...
(b) Find the sum of the first 12 terms of the following geometric sequence:
8, 4, 2, 1,...
(a) Step 1: To find the sum we identify the following:
First term a = -3
The common ratio, r = -2 (each term is the previous term multiplied by -2)
The number of terms, n = 6.
Step 2: Apply the sum of Geometric Sequence formula to the information in Step 1 to get:
S = a(r^n - 1)/(r - 1) = -3((-2)^6 - 1)/(-2 - 1) = -3(63)/(-3) = 63.
The answer is 63.
(b) Step 1: To find the sum we identify the following:
The first term, a = 8
The common ratio, r = 1/2 = 0.5 (each term is the previous term multiplied by 1/2)
The number of terms, n = 12.
Step 2: Apply the sum of Geometric Sequence formula to the information in Step 1 to get:
S = a(r^n - 1)/(r - 1) = 8((0.5)^(12) - 1)/(0.5 - 1) = 8(-0.9997)/(-0.5) = 15.9960.
The answer is 15.9960.
A geometric sequence (or geometric progression) is a sequence of numbers in which each one is the product of the previous number and a constant, known as the common ratio. A sum of a geometric progression is the sum of a given number of terms from the sequence, in order.
A series is the sum of the terms of a well defined sequence. A geometric sequence is one in which each term is the product of the previous term and a given constant, and a geometric series is a series based on a geometric sequence.
Where a is the initial value and r is the common ratio: a * (1/(1 - r))
Note that this formula only applies if |r|<1! If this is not the case, then the series diverges.
Where a is the initial value and r is the common ratio:
a * (1 - r^(n+1))/(1 - r)
Remember that to find a given term, n should be one less than the number of that term, as the index begins at 0. For instance, to find the 10th term, n=9.
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