How can a finite geometric sum be extended to an infinite geometric series? In what circumstances can we quickly find the value of an infinite geometric series?
In our work in Section 8.2, we learned how to find a degree \(n\) polynomial approximation centered at a value \(a\) for a given function \(f\) with at least \(n\) derivatives. By working with several different functions and \(n\)-values, we’ve seen that increasing the degree of the polynomial improves the approximation, and also often helps us to see a pattern in the coefficients of the Taylor polynomials.
and we see a pattern in the coefficients that allows us to easily generate \(T_6(x)\text{,}\)\(T_{10}(x)\text{,}\) or indeed \(T_n(x)\) for any \(n\text{.}\) Note that if we want to use \(T_{10}(x)\) to estimate \(\ln(\frac{3}{2}) = \ln(1 + \frac{1}{2})\text{,}\) we need to compute the sum of \(10\) terms given by
Observe that we can think of the first term, \(1\text{,}\) as \((\frac{1}{2})^0\text{,}\) and since the powers of \(\frac{1}{2}\) include every whole number from \(0\) to \(n-1\text{,}\) there are exactly \(n\) terms in the sum given by \(S_n\text{.}\)
Note that \(S_1 = 1\text{,}\)\(S_2 = S_1 + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}\text{,}\) and \(S_3 = S_2 + \frac{1}{4} = \frac{3}{2} + \frac{1}{4} = \frac{7}{4}\text{.}\) Using the fact that each subsequent value of \(S_n\) can be computed by adding one additional term to the preceding sum, complete Table 8.3.1 with the exact (fractional) value of each sum.
all share a similar structure: each has exactly \(n\) terms, and the next term in each sum is found by multiplying the last term by the same number. In the first sum, each subsequent term is found by multiplying by \(\frac{1}{2}\text{;}\) in the second sum, by multiplying by \(\frac{1}{4}\text{;}\) in the third sum, the multiplier is \(\frac{-2}{3}\text{.}\) These sums each have the form
which we call a finite geometric series with ratio \(r\text{.}\) It turns out that the value of each sum that has this form can be computed quickly without having to add all of the individual terms.
Our goal in this activity is to find a shortcut formula for \(S_n\) that can be written as a single fraction that does not involve a sum of \(n\) terms.
The ideas in Activity 8.3.2 can be extended to the general case for any value of \(a\) and any value of \(r \ne 1\text{.}\) In particular, replacing \(1\) with \(a\) and \(\frac{2}{5}\) with \(r\text{,}\) our work shows that by finding \(r \cdot S_n\) and then subtracting that quantity from \(S_n\text{,}\) we get
\begin{equation*}
S_n - rS_n = (a + ar + \cdots + ar^{n-2} + ar^{n-1}) - (ar + ar^2 + \cdots + ar^{n-1} + ar^{n})\text{,}
\end{equation*}
so that
\begin{equation*}
S_n(1-r) = a - ar^{n}\text{,}
\end{equation*}
A finite geometric series \(S_n\) is a sum of the form
\begin{equation}
S_n = a + ar + ar^2 + \cdots + ar^{n-1}\text{,}\tag{8.3.6}
\end{equation}
where \(a\) and \(r\) are real numbers such that \(r \ne 1\text{.}\) The exact value of the finite geometric series \(S_n\) can be computed directly as
We can view the given sum as the finite geometric series \(S_{10}\) that has \(a = 1\) and \(r = \frac{1}{2}\text{.}\) By Equation (8.3.7), it follows that
We can view the given sum as the finite geometric series \(S_{8}\) that has \(a = 2\) and \(r = -\frac{4}{3}\text{.}\) By Equation (8.3.7), it follows that
Note that \(6561 = 3^8\text{,}\) so since this sum begins with \(\frac{1}{3} = 3^{-1}\text{,}\) this is a finite geometric series with \(10\) terms. Hence we can view the given sum as the finite geometric series \(S_{10}\) that has \(a = \frac{1}{3}\) and \(r = 3\text{:}\)
Because the approximation gets better as we add more terms, it’s natural to think about the possibility of the sum extending forever. We begin by asking this question for a finite geometric series such as
whose value we seek. Plotting these partial sums on a number line, we see evidence that the value of the \(n\)th partial sum is approaching \(2\text{.}\)
Figure8.3.6.A plot of \(S_1\text{,}\)\(S_2\text{,}\)\(S_3\text{,}\)\(S_4\text{,}\)\(S_5\text{,}\) and the number \(2\) on the interval \([1,2]\text{.}\)
Indeed, we observe that each partial sum lies halfway between the preceding partial sum and the number \(2\text{:}\)\(S_2 = 1.5\) is halfway between \(S_1 = 1\) and \(2\text{;}\)\(S_3 = 1.75\) is halfway between \(S_2 = 1.5\) and \(2\text{;}\) and so on. This shows that the partial sums \(S_n\) are approaching \(2\) as \(n\) increases without bound.
We can see this more formally in Equation (8.3.8) if we divide the two terms in the numerator of \(S_n\) by the denominator. Doing so, an equivalent formula for \(S_n\) is
Example 8.3.5 demonstrates the general principle that we use to determine if any infinite series has a finite value: we consider the partial sum, \(S_n\text{,}\) which is the finite sum of the first \(n\) terms, and then investigate whether the partial sums converge to a single value as \(n\) increases without bound. For geometric series, determining whether the partial sums converge or not is straightforward.
Equation (8.3.11) holds because when \(|r| \lt 1\text{,}\) it follows that \(r^{n} \to 0\) as \(n \to \infty\text{.}\) Indeed, when \(|r| \lt 1\text{,}\) we see by taking the limit as \(n \to \infty\) in Equation (8.3.7) that
As in our work in Chapter 4 with Riemann sums, we can use sigma notation (see also Subsection 4.2.2) to express a sum in convenient shorthand. For instance, it’s equivalent to write
\begin{equation*}
S_n = a + ar + ar^2 + \cdots + ar^{n-1} = \sum_{k=0}^{n-1} ar^{k} \text{.}
\end{equation*}
For each of the following infinite geometric series, determine the values of \(a\) and \(r\text{,}\) compute the partial sums \(S_{5}\) and \(S_{10}\) exactly (writing each as a fraction), and if the infinite geometric series converges, find its value.
provided that \(|r| \lt 1\text{.}\) To study this equation further, we are going to let \(r\) vary and thus we introduce the function \(f(x) = \frac{1}{1-x}\) and replace \(r\) by \(x\) in Equation (8.3.12) to have
which is valid for \(|x| \lt 1\text{.}\) One reason this equation is interesting is that we have a function \(f(x)\) that can be represented in two different ways: as the rational function \(\frac{1}{1-x}\text{,}\) and as the infinite polynomial function \(1 + x + x^2 + \cdots\text{.}\)
For \(x\) such that \(|x| \lt 1\text{,}\) we know from Equation (8.3.13) that the infinite geometric series \(1 + x+ x^2 + \cdots + x^{n-1} + x^n + \cdots\) converges. Thus, it follows that partial sums of the series will approximate its value, which means
Since the series converges for \(|x| \lt 1\text{,}\) this also means the approximate equality in (8.3.14) holds for \(x\) near \(a = 0\text{.}\) This result reminds us of approximations generated by Taylor polynomials of degree \(n\text{.}\)
Moreover, because the infinite series converges, the larger the value of \(n\text{,}\) the better the approximation will be. Plotting the function \(f(x) = \frac{1}{1-x}\) along with several of the polynomials that arise for different choices of \(n\text{,}\) say \(P_1(x) = 1 + x\text{,}\)\(P_4(x) = 1 + x + \cdots + x^4\text{,}\) and \(P_7(x) = 1 + x + \cdots + x^7\text{,}\) we see the impact of increasing the degree \(n\) in Figure 8.3.7.
In particular, we observe that as the degree of the polynomial approximation increases, the polynomial not only appears closer to \(f(x) = \frac{1}{1-x}\text{,}\) but does so on a wider interval of \(x\)-values. Since the infinite series only converges when \(|x| \lt 1\) and the function \(f(x) = \frac{1}{1-x}\) is undefined when \(x = 1\text{,}\) we also expect the approximations to only be accurate on an interval that lies within \(-1 \lt x \lt 1\text{.}\) This also reminds of our earlier work with Taylor polynomials where in images such as Figure 8.2.7, increasing the degree of the Taylor polynomial similarly improves the approximation.
and exploring its partial sums. Next, we change perspective and start with function \(f(x) = \frac{1}{1-x}\) and determine the Taylor polynomial approximations to \(f\) that are centered at \(a = 0\) in order to see an interesting connection.
To begin finding \(T_n(x)\text{,}\) do the usual work of computing the various derivatives of \(f\) and their respective values at \(a = 0\text{;}\) note that it’s helpful to view \(f(x)\) in the form \(f(x) = (1-x)^{-1}\) so that we can easily compute the derivatives of \(f\) using the chain rule. For instance,
where the first “\(-1\)” arises from the power rule, while the second “\(-1\)” results from the chain rule, since \(\frac{d}{dx}[1-x] = -1\text{.}\) In order to see key patterns that arise, it’s helpful not to combine the products of numbers that arise in the various derivatives. Record the first five derivatives of \(f(x)\) in the spaces provided below.
Next, evaluate the derivatives you determined in (a) at \(a = 0\) and use these to find the values of the coefficients of the Taylor polynomial centered at \(a = 0\text{.}\) Record your work in the blank spaces provided.
What pattern do you observe in the value of \(c_k\text{?}\) State the degree \(5\) Taylor polynomial, \(T_5(x)\text{,}\) as well as the formula you expect for the general degree \(n\) Taylor polynomial, \(T_n(x)\text{.}\)
What is the Taylor series centered at \(a = 0\) for \(f(x) = \frac{1}{1-x}\text{?}\) (As we will see in the next section, the “Taylor series” of a function is the infinite series that results by simply extending the Taylor polynomials indefinitely.)
For each of the finite geometric series given below, indicate the number of terms in the sum and find the sum. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.
Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.
What do you observe happens to these partial sums as \(n\) increases without bound? What does this tell us about the infinite geometric series \(1 + 2 + 4 + \cdots + 2^{n-1} + \cdots\text{?}\)
Suppose you drop a golf ball onto a hard surface from a height \(h\text{.}\) The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height \(\frac{3}{4}\) of the height from which it dropped. Let \(h_n\) be the height of the ball on the \(n\)th bounce, with \(h_0 = h\text{.}\) In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance.
Determine a formula for \(h_1\) in terms of \(h\text{.}\)
Next, let’s determine the total amount of time the ball is in the air.
When the ball is dropped from a height \(H\text{,}\) if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time \(t\) is given by
\begin{equation*}
H - \frac{1}{2}gt^2\text{.}
\end{equation*}
Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height \(H\text{.}\)
It is important to understand the power of geometric growth compared to linear growth. Suppose you are hired for a job that will take you 30 days to complete and are offered two options for how you’ll be compensated.
You can be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on, doubling the amount you are paid each day.
Find a formula for the amount paid on day \(n\text{,}\) as well as for the total amount paid by day \(n\text{.}\) Use this formula to determine which option (1 or 2) you should take.
