9.1: Sequences and Series

In the 5^th century B.C. the ancient Greek philosopher Zeno of Elea devised several paradoxes, the most famous of which—The Dichotomy—asserts that if space is infinitely divisible then motion is impossible. The argument goes like this: imagine a line segment of finite length, say 1 m, with a person at one end as in Figure [fig:zeno].

Before traversing the entire distance the person would first need to travel one half the distance. Before doing that, though, he would need to travel one fourth the distance, and before that one eighth the distance, and so on. There is thus no “first” distance for him to traverse, meaning his motion cannot even begin!

Obviously motion is possible, or you would not be reading this. Does that mean Zeno’s reasoning is flawed? More on that later. In the meantime, notice a few things in Figure [fig:zeno]. First, the distance markers \(\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots\) form an infinite sequence of numbers approaching 0. Second, the sum of the distances between successive markers is an infinite series which should equal the total segment length 1:

\[\frac{1}{2} ~+~ \frac{1}{4} ~+~ \frac{1}{8} ~+~ \cdots ~=~ 1 \nonumber \]

It will be shown shortly that the sum is indeed 1, which turns out to have no bearing on Zeno’s paradox. First some definitions are needed. A sequence is an ordered list of objects, which in this book will always be real numbers. Sequences can be finite or infinite: finite if there is a last number in the list, infinite if every number in the list is followed by another number (i.e. a “successor”). Sequences should not be confused with sets—order matters in a sequence but not in a set, and numbers may repeat in a sequence but not in a set.

For example, the sequences \(\langle 1,2,3\rangle\) and \(\langle 1,3,2\rangle\) are different, since order matters. However, the numbers in those sequences comprise the same set \(\lbrace 1,2,3\rbrace\).

The simplest example of an infinite sequence is \(\Naturals\), the set of natural numbers: \(0,1,2,3,\ldots\). In fact, the numbers \(a_0,a_1,a_2,\ldots\) in any infinite sequence of real numbers can be written as the range of a function \(f\) mapping \(\Naturals\) into \(\Reals\):

\[\label{eqn:seqfcn} f(n) ~=~ a_n \]

Typically notation such as \(\seq{a_n}_{n=0}^{\infty}\) is used for representing an infinite sequence, or simply \(\seq{a_n}\) when the initial value of the index \(n\) is understood (and \(n\) is always an integer). The intuitive notion of the limit of an infinite sequence can be stated formally:

In other words, a sequence \(\seq{a_n}\) converges to \(L\) if the terms \(a_n\) can be made arbitrarily close to \(L\) for \(n\) sufficiently large. In most cases the formal definition will not be needed, since by formula ([eqn:seqfcn]) the same rules and formulas from Chapters 1 and 3 for the limit of a function \(f(x)\) as \(x\) approaches \(\infty\) apply to sequences (e.g. sums and products of limits, L’Hôpital’s Rule). All you have to do is replace \(x\) by \(n\).

For integers \(n \ge 1\) define \(a_n = \frac{1}{2^n}\). Find \(\displaystyle\lim_{n \to \infty} a_n\) if it exists.

Solution: Since \(\displaystyle\lim_{x \to \infty} \,\frac{1}{2^x} = 0\) then replacing \(x\) by \(n\) shows that

\[\lim_{n \to \infty} a_n ~=~ \lim_{n \to \infty}~ \frac{1}{2^n} ~=~ 0 ~. \nonumber \]

For integers \(n \ge 0\) define \(a_n = \frac{2n+1}{3n+2}\). Is \(\seq{a_n}\) a convergent sequence? If so then find its limit.

Solution: By L’Hôpital’s Rule, treating an integer \(n\ge 0\) as a real-valued variable \(x\),

\[\lim_{n \to \infty} a_n ~=~ \lim_{n \to \infty}~ \frac{2n+1}{3n+2} ~=~ \lim_{n \to \infty}~ \frac{2}{3} ~=~ \frac{2}{3} ~. \nonumber \]

Thus the sequence is convergent and its limit is \(\frac{2}{3}\).

For integers \(n \ge 0\) define \(a_n = \frac{e^n}{3n+2}\). Is \(\seq{a_n}\) a convergent sequence? If so then find its limit.

Solution: By L’Hôpital’s Rule, treating an integer \(n\ge 0\) as a real-valued variable \(x\),

\[\lim_{n \to \infty} a_n ~=~ \lim_{n \to \infty}~ \frac{e^n}{3n+2} ~=~ \lim_{n \to \infty}~ \frac{e^n}{3} ~=~ \infty \nonumber \]

Thus the sequence is divergent.

Example \(\PageIndex{1}\): fib

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Solution

The famous Fibonacci sequence¹ \(\seq{F_n}\) starts with the numbers 0 and 1, then each successive term is the sum of the previous two terms: \(F_0 = 0\), \(F_1 = 1\),

\[\label{eqn:fibonacci} F_n ~=~ F_{n-1} ~+~ F_{n-2} ~~\text{for integers $n \ge 2$} \]

Equation ([eqn:fibonacci]) is a recurrence relation. The first ten Fibonacci numbers are \(0,1,1,2,3,5,8,13,21,34\). Clearly \(\seq{F_n}\) is a divergent sequence, since \(F_n \rightarrow \infty\). For \(n \ge 2\) define \(a_n = F_n/F_{n-1}\). The first few values are:

\[a_2 ~=~ \frac{F_2}{F_1} ~=~ \frac{1}{1} ~=~ 1 \quad,\quad a_3 ~=~ \frac{F_3}{F_2} ~=~ \frac{2}{1} ~=~ 2 \quad,\quad a_4 ~=~ \frac{F_4}{F_3} ~=~ \frac{3}{2} ~=~ 1.5 \quad,\quad a_5 ~=~ \frac{F_5}{F_4} ~=~ \frac{5}{3} ~\approx~ 1.667 \nonumber \]

Show that \(\seq{a_n}\) is convergent. In other words, in the Fibonacci sequence the ratios of each term to the previous term converge to some number.

Solution: There are many ways to prove this, perhaps the simplest being to assume the sequence is convergent and then find the limit (which would be impossible if the sequence were divergent). So assume that \(a_n \rightarrow a\) for some real number \(a\). Then divide both sides of formula ([eqn:fibonacci]) by \(F_{n-1}\), so that

\[\frac{F_n}{F_{n-1}} ~=~ 1 ~+~ \frac{F_{n-2}}{F_{n-1}} \quad\Rightarrow\quad a_n ~=~ 1 ~+~ \frac{1}{a_{n-1}} ~~\text{for integers $n \ge 2$} \nonumber \]

Now take the limit of both sides of the last equation as \(n \to \infty\):

\[\lim_{n \to \infty} ~a_n ~=~ \lim_{n \to \infty} ~\left(1 ~+~ \frac{1}{a_{n-1}}\right) \quad\Rightarrow\quad a ~=~ 1 ~+~ \frac{1}{a} \quad\Rightarrow\quad a^2 - a - 1 ~=~ 0 \quad\Rightarrow\quad a ~=~ \frac{1 \pm \sqrt{5}}{2} \nonumber \]

Since \(a\) must be positive then \(a = \frac{1 +\sqrt{5}}{2}\). Thus, the sequence is convergent and converges to \(\frac{1 +\sqrt{5}}{2} \approx 1.618\).
Note: This number is the famous golden ratio, the subject of many claims regarding its appearance in nature and supposed aesthetic appeal as a ratio of sides of a rectangle.²

An infinite series is the sum of an infinite sequence. If the infinite sequence is \(\seq{a_n}_{n=0}^{\infty}\) then the series can be written as

\[\sum_{n=0}^{\infty} a_n ~=~ a_0 + a_1 + a_2 + \cdots + a_n + \cdots \nonumber \]

or simply as \(\sum a_n\) when the initial value of the index \(n\) is understood. There is a natural way to define the sum of such a series:

One important convergent series is a geometric progression:

\[\label{eqn:geomprog} a + ar + ar^2 + ar^3 + \cdots + ar^n + \cdots \]

with \(a \ne 0\) and \(\abs{r} < 1\). Multiply the \(n\)-th partial sum \(s_n\) by \(r\):

\[rs_n ~=~ r\,(a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n) ~=~ ar + ar^2 + ar^3 + ar^4 + \cdots + ar^n + ar^{n+1} \nonumber \]

Now subtract \(rs_n\) from \(s_n\):

\[\begin{aligned} s_n - rs_n ~&=~ a + ar + ar^2 + ar^3 + \cdots + ar^n - (ar + ar^2 + ar^3 + ar^4 + \cdots + ar^n + ar^{n+1})\\ s_n - rs_n ~&=~ a - ar^{n+1} \quad\text{so that}\\ \lim_{n \to \infty} s_n ~&=~ \lim_{n \to \infty} \frac{a\,(1 - r^{n+1})}{1 -r} ~=~ \frac{a}{1-r}\end{aligned} \nonumber \]

since \(r^{n+1} \rightarrow 0\) as \(n \to \infty\) when \(\abs{r} < 1\). Thus:

Example \(\PageIndex{1}\): geomzeno

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Solution

Show that \(\displaystyle\sum_{n=0}^{\infty} \,\frac{1}{2^n} = 1\).

Solution: This is a geometric progression with \(a=\frac{1}{2}\) and \(r=\frac{1}{2}\). So by formula ([eqn:geomprogsum]) the sum is:

\[\sum_{n=0}^{\infty} \;\frac{1}{2^n} ~=~ \frac{a}{1-r} ~=~ \frac{\frac{1}{2}}{1 - \frac{1}{2}} ~=~ 1 \quad\checkmark \nonumber \]

Write the repeating decimal \(0.\overline{17} = 0.17171717\ldots\) as a rational number.

Solution: This is a geometric progression with \(a=0.17=\frac{17}{100}\) and \(r=0.01=\frac{1}{100}\):

\[\begin{aligned} 0.171717\ldots ~&=~ 0.17 + 0.0017 + 0.000017 + \cdots ~=~ 0.17\,(0.01)^0 + 0.17\,(0.01)^1 + 0.17\,(0.01)^2 + \cdots\\ &=~ \sum_{n=0}^{\infty} \;0.17\,(0.01)^n ~=~ \frac{a}{1-r} ~=~ \frac{0.17}{1 - 0.01} ~=~ \frac{0.17}{0.99} ~=~ \frac{17}{99}\end{aligned} \nonumber \]

Back to Zeno’s motion paradox, by Example

Example \(\PageIndex{1}\): geomzeno

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Solution

the sum of the infinite number of distances between successive markers in Figure [fig:zeno] is 1, as expected. This fact is often mistaken as proof that Zeno was wrong, yet it does not actually address Zeno’s argument, since the person is attempting to begin motion at the tail end—the “infinite end”—of the geometric progression, not at the beginning (i.e. at \(n=0\)). Zeno’s point remains that there is no “first step” at that tail end.

In fact, even if you were to reverse the person’s position to start at the other end, so that he would first move a distance \(\frac{1}{2}\), then a distance \(\frac{1}{4}\), and so on, as in Figure [fig:zenoreverse], then a new problem is introduced: the person keeps moving no matter how close he is to the end point. There is now no “last step” and motion is thus still impossible.

The convergence of the geometric progression has misled many people to argue that Zeno is wrong, by claiming it shows an infinite number of movements can be completed in a finite amount of time. But this line of argument fails on (at least) two counts. First, Zeno never argued about time—it is irrelevant to his paradox. The more fundamental flaw is that the introduction of time brings along the concept of speed, typically taken to be constant (though it need not be). Speed is distance over time, but it is precisely the distance part of that ratio that Zeno rejects—that distance can never be traveled. In other words, it is circular—and hence faulty—reasoning to “prove” that motion is possible by assuming it is possible.

The geometric progression also doesn’t help when considering only the distances and ignoring time. Even assuming each of those distances could be traveled, the partial sums approach 1 but never actually reach it. The limit of a sequence is defined in terms of an inequality—you can get arbitrarily close to the limit, and that is all. The equality in formula ([eqn:geomprogsum]) is merely a shorthand way of saying that. It is an abstraction based on properties of the real number system, not necessarily based on physical reality.

Zeno’s paradox is not purely mathematical—it is about space and hence is physical, with a tinge of philosophy. Physicists have recognized this—and noted the flaw in the purely mathematical line of attack—and devised new arguments against Zeno, some more sophisticated than others. However, they all invariably end up in some sort of circular reasoning. All this calls into question the original assumption: the infinite divisibility of space. If space had some smallest unit that could not be divided any further then there is no paradox—motion over a finite distance could always be decomposed into a large but finite number of irreducible steps.³

[sec9dot1]

For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit.

4

\(\seq{n\,e^{-n}}_{n = 0}^{\infty}\vphantom{\seq{\dfrac{n^2}{3n^2 ~+~ 7n ~-~ 2}}_{n = 1}^{\infty}}\)

\(\seq{\dfrac{n^2}{3n^2 ~+~ 7n ~-~ 2}}_{n = 1}^{\infty}\)

\(\seq{\dfrac{n^2}{3n^3 ~+~ 7n ~-~ 2}}_{n = 1}^{\infty}\)

\(\seq{\dfrac{n^3}{3n^2 ~+~ 7n ~-~ 2}}_{n = 1}^{\infty}\)

4

\(\seq{\dfrac{n}{\ln\,n}}_{n = 2}^{\infty}\vphantom{\seq{\dfrac{n\,!}{(n + 2)\,!}}_{n = 0}^{\infty}}\)

\(\seq{\dfrac{n\,!}{(n + 2)\,!}}_{n = 0}^{\infty}\)

\(\seq{\sin\,\left(\dfrac{n \pi}{2}\right)}_{n = 0}^{\infty}\vphantom{\seq{\dfrac{n\,!}{(n + 2)\,!}}_{n = 0}^{\infty}}\)

\(\seq{(-1)^n \cos\,n\pi}_{n = 0}^{\infty}\vphantom{\seq{\dfrac{n\,!}{(n + 2)\,!}}_{n = 0}^{\infty}}\)

For Exercises 9-12 determine whether the given series is convergent. If so then find its sum. [[1.]]

4

\(\bigsum{n = 0}{\infty}~ 2\;\left(\dfrac{2}{3}\right)^n\)

\(\bigsum{n = 1}{\infty}~ 7^{-n}\)

\(\bigsum{n = 0}{\infty}~ \dfrac{3^n ~+~ 5^{n + 1}}{6^n}\)

\(\bigsum{n = 0}{\infty}~ n\)

For Exercises 13-16 use a geometric progression to write the given repeating decimal as a rational number. [[1.]]

4

\(0.\overline{113}\)

\(0.\overline{9}\)

\(0.24\overline{9}\)

\(0.01\overline{7}\)

[[1.]]

In Example

Example \(\PageIndex{1}\): fib

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Solution

define \(b_n = F_n/F_{n+1}\) for \(n\ge 0\) and show that \(\seq{b_n}_{n=0}^{\infty}\) converges to \(\frac{\sqrt{5}-1}{2}\).

Show that formula ([eqn:newton]) for Newton’s method with \(f(x) = x^2 - 2\) and \(x_0 = 1\) yields the sequence \(\seq{x_n}_{n=0}^{\infty}\) with

\[x_n ~=~ \frac{1}{2} x_{n-1} ~+~ \frac{1}{x_{n-1}} \quad\text{for $n\ge 1\;$,} \nonumber \]

then assuming \(\seq{x_n}\) is convergent show that it must converge to \(\sqrt{2}\). (Hint: See Example

Example \(\PageIndex{1}\): fib

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Solution

.)

In this exercise you will prove a formula for the general number \(F_n\) in the Fibonacci sequence. Denote the positive and negative solutions to the equation \(x^2 - x - 1 = 0\) by \(\phi_{+} = \frac{1+\sqrt{5}}{2}\) and \(\phi_{-} = \frac{1-\sqrt{5}}{2}\), respectively. Note that \(\phi_{+}\) is the golden ratio mentioned in Example

Example \(\PageIndex{1}\): fib

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Solution

.

1. Use the equation \(x^2 - x - 1 = 0\) to show that

   \[\phi_{+}^{n+1} ~=~ \phi_{+}^{n} ~+~ \phi_{+}^{n-1} \qquad\text{and}\qquad \phi_{-}^{n+1} ~=~ \phi_{-}^{n} ~+~ \phi_{-}^{n-1} ~. \nonumber \]

2. Use part (a) and induction to show that

   \[F_n ~=~ \frac{\phi_{+}^{n} \;-\; \phi_{-}^{n}}{\sqrt{5}} \quad\text{for $n\ge 0$.} \nonumber \]

A ball is dropped from a height of 4 ft above the ground, and upon each bounce off the ground the ball bounces straight up to a height equal to 65% of its previous height. Find the theoretical total distance the ball could travel if it could bounce indefinitely. Why is this physically unrealistic?

A continued fraction is a type of infinite sum involving a fraction with a denominator that continues indefinitely:

\[a ~=~ a_0 ~+~ \cfrac{1}{a_1 \;+\; \cfrac{1}{a_2 \;+\; \cfrac{1}{a_3 \;+\; \cdots}}} \nonumber \]

1. Show that the golden ratio \(\phi_{+} = \frac{1+\sqrt{5}}{2}\) (a solution of \(x^2-x-1=0\)) can be written as

   \[\phi_{+} ~=~ 1 ~+~ \cfrac{1}{1 \;+\; \cfrac{1}{1 \;+\; \cfrac{1}{1 \;+\; \cdots}}} ~. \nonumber \]

   (Hint: Look for a recurrence relation in the fraction.)

2. Show that

   \[\sqrt{2} ~=~ 1 ~+~ \cfrac{1}{2 \;+\; \cfrac{1}{2 \;+\; \cfrac{1}{2 \;+\; \cdots}}} ~. \nonumber \]

[exer:goldensqrt] Show that the golden ratio \(\phi_{+} = \frac{1+\sqrt{5}}{2}\) can be written as \(\phi_{+} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}\).

Write a computer program to approximate the result in Exercise [exer:goldensqrt] using \(100\) terms. After how many iterations do the approximate values start repeating?

Would the existence of infinitesimals as a measure of space resolve Zeno’s paradox? Explain.

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1. Due to the Italian mathematician Leonardo Fibonacci (ca. 1170-1250).↩
2. For example, see Huntley, H.E., The Divine Proportion, New York: Dover Publications, Inc., 1970.↩
3. As of this writing there is not yet a definitive answer as to whether space is continuous or discrete (quantized). A “smallest unit” would have to be below the Planck level—around \(10^{-33}\) cm, well below current measurement capabilities. Some recent advances in the field of loop quantum gravity suggest the possibility of quantized space. See Chamseddine, A.H., Connes, A. & Mukhanov, V., “Geometry and the quantum: basics.” J. High Energy Phys. 2014, 98 (2014). https://doi.org/10.1007/JHEP12(2014)098↩
4. For a proof see pp.48-49 in Buck, R.C., Advanced Calculus, 2nd ed., New York: McGraw-Hill Book Co., 1965.↩
5. See pp.612-613 in Taylor, A.E. and W.R. Mann, Advanced Calculus, 2nd ed., New York: John Wiley & Sons, Inc., 1972.↩
6. Due to the English mathematician and theologian John Wallis (1616-1703). For a proof of the formula see pp.738-739 in Taylor, A.E. and W.R. Mann, Advanced Calculus, 2nd ed., New York: John Wiley & Sons, Inc., 1972.↩
7. For formal proofs of these statements, see pp.442-444 in Klambauer, G., Aspects of Calculus, New York: Springer-Verlag, 1986.↩
8. The symbol \(\aleph_0\) is called aleph null and represents the cardinality of \(\Naturals\), i.e. its size. \(\aleph_0\) is the smallest infinity.↩
9. The proofs require uniform convergence, a stronger condition than ordinary convergence. See pp.129-134 in Bromwich, T.J., An Introduction to the Theory of Infinite Series, 2nd ed., London: Macmillan & Co. Ltd., 1955.↩
10. Due to the German astronomer Friedrich Wilhelm Bessel (1784-1846), from a study of elliptic planetary motion.↩
11. Named after English mathematician Brook Taylor (1685-1731), though such series were known to others (e.g. James Gregory, Johann Bernoulli) before Taylor.↩
12. The special case of \(c=0\) in Taylor’s formula yields what is sometimes called the Maclaurin’s series for \(f(x)\), though that terminology is typically not used in fields of study outside mathematics.↩
13. To see how inaccurate calculators can be, compute \(\tan (355/226)\) in radian mode on a calculator. The true value to 3 decimal places is \(-7497258.185\), but few calculators produce an answer close to that.↩