Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Harmonic sequence

Revision as of 20:43, 26 November 2021 by Etmetalakret (talk | contribs)

In algebra, a harmonic sequence, sometimes called a harmonic progression, is a sequence of numbers such that the difference between the reciprocals of any two consecutive terms is constant. In other words, a harmonic sequence is formed by taking the reciprocals of every term in an arithmetic sequence.

For example, $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{99}, \textrm{ } \frac{1}{91}, \textrm{ } \frac{1}{83}, \frac{1}{75}$ are harmonic sequences; however, $1, 1, \frac{1}{3}, \frac{1}{5}$ and $\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots$ are not. By definition, $0$ can never be a term of a harmonic sequence.

More formally, a harmonic progression $a_1, a_2, \ldots , a_n$ biconditionally satisfies $1/a_2 - 1/a_1 = 1/a_3 - 1/a_2 = \cdots = 1/a_n - 1/a_{n-1}.$ A similar definition holds for infinite harmonic sequences. It appears most frequently in its three-term form: namely, that constants $a$, $b$, and $c$ are in harmonic progression if and only if $1/b - 1/a = 1/c - 1/b$.

Properties

Because the reciprocals of the terms in a harmonic sequence are in arithmetic progression, one can apply properties of arithmetic sequences to derive a general form for harmonic sequences. Namely, for some constants $a$ and $d$, the terms of any harmonic sequence can be written as \[\frac{1}{a}, \textrm{ } \frac{1}{a+d}, \textrm{ } \frac{1}{a+2d}, \textrm{ } \cdots \textrm{ } \frac{1}{a+(n-1)d}.\]

A common lemma is that a sequence is in harmonic progression if and only if $a_n$ is the harmonic mean of $a_{n-1}$ and $a_{n+1}$ for any consecutive terms $a_{n-1}, a_n, a_{n+1}$. In symbols, $2/a_n = 1/a_{n-1} + 1/a_{n+1}$. This is mostly used to perform substitutions, though it occasionally serves as a definition of harmonic sequences.

Sum

A harmonic series is the sum of all the terms in a harmonic series. All infinite harmonic series diverges; this is by a limit comparison test with the series $1 + 1/2 + 1/3 + \cdots$, which is referred to as the harmonic series. As for finite harmonic series, a general expression for the sum has ever been found. One must find a strategy to evaluate their sum on a case-by-case basis.

Examples

Here are some example solutions that utilize harmonic sequences and series.

Example 1

Find all real numbers such that $x+4, x+1, x$ is a harmonic sequence.

Solution: Using the harmonic mean properties of harmonic sequences, \[\frac{2}{x+1} = \frac{1}{x+4} + \frac{1}{x} = \frac{2x+4}{x(x+4)}.\] Note that $x=-4, -1, 0$ would create a term of $0$—something that breaks the definition of harmonic sequences—which eliminates them as possible solutions. We can thus multiply both sides by $x(x+1)(x+4)$ to get $2x(x+4) = (2x+4)(x+1)$. Expanding these factors yields $2x^2 + 8x = 2x^2 + 6x + 4$, which simplifies to $x=2$. Thus, $2, 3, 6$ is the only solution to the equation, as desired. $\square$

Example 2

Let $a$, $b$, and $c$ be positive real numbers. Show that if $a$, $b$, and $c$ are in harmonic progression, then $a/(b+c)$, $b/(c+a)$, and $c/(a+b)$ are as well.

Solution: Using the harmonic mean property of harmonic sequences, we are given that $1/a + 1/b = 2/c$, and we wish to show that $(b+c)/a + (a+b)/c = 2(c+a)/b$. We work backwards from the latter equation.

One approach might be to add $2$ to both sides of the equation, which yields \[\frac{a+b+c}{a} + \frac{a+b+c}{c} = \frac{2(a+b+c)}{b}.\] Because $a$, $b$, and $c$ were all defined to be positive, $a+b+c \neq 0$. Thus, we can divide both sides of the equation by $a+b+c$ to get $1/a + 1/c = 2/b$, which was given as true.

From here, it is easy to write the proof forwards. Then $(b+c)/a + (a+b)/c = 2(c+a)/b$, which implies that the sequence $a/(b+c), b/(c+a), c(a+b)$ is in harmonic progression, as required. $\square$

Example 3

2019 AMC 10A Problem 15: A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \frac{3}{7}$, and $a_n = \frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2} - a_{n-1}}$ for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $p+q$?

Solution: We simplify the series recursive formula. Taking the reciprocals of both sides, we get the equality \[\frac{1}{a_n} = \frac{2 a_{n-2} - a_{n-1}}{a_{n-1} \cdot a_{n-2}} = \frac{2}{a_{n-2}} - \frac{1}{a_{n-1}}.\] Thus, $2/a_{n_1} = 1/a_{n-2} + 1/a_n$. By an above lemma, the entire sequence is in harmonic progression, which means that we can apply tools of harmonic sequences to this problem.

We will now find a closed expression for the sequence. Let $a_1 = 1/a$ and $a_2 = 1/(a+d)$. Simplifying the first equation yields $a=1$ and substituting this into the second equation yields $d = 4/3$. Thus, \[a_n = \frac{1}{1 + \frac{4}{3}(n-1)},\] and so $a_{2019} = 8075 / 3$. The answer is then $8075 + 3 = 8078$, or $E$. $\square$

More Problems

Here are some more problems that utilize harmonic sequences and series. Note that harmonic sequences are rather uncommon compared to their arithmetic and geometric counterparts.

Introductory

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Harmonic_sequence&oldid=166701"