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Difference between revisions of "Harmonic sequence"

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We will now find a closed expression for the sequence. Let <math>a_1 = 1/a</math> and <math>a_2 = 1/(a+d)</math>. Simplifying the first equation yields <math>a=1</math> and substituting this into the second equation yields <math>d = 4/3</math>. Thus, <cmath>a_n = \frac{1}{1 + \frac{4}{3}(n-1)},</cmath> and so <math>a_{2019} = 8075 / 3</math>. The answer is then <math>8075 + 3 = 8078</math>, or <math>E</math>. <math>\square</math>
 
We will now find a closed expression for the sequence. Let <math>a_1 = 1/a</math> and <math>a_2 = 1/(a+d)</math>. Simplifying the first equation yields <math>a=1</math> and substituting this into the second equation yields <math>d = 4/3</math>. Thus, <cmath>a_n = \frac{1}{1 + \frac{4}{3}(n-1)},</cmath> and so <math>a_{2019} = 8075 / 3</math>. The answer is then <math>8075 + 3 = 8078</math>, or <math>E</math>. <math>\square</math>
  
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=== More Problems ===
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== More Problems ==
 
Here are some problems that utilize harmonic sequences and series.
 
Here are some problems that utilize harmonic sequences and series.
  

Revision as of 14:39, 26 November 2021

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 $2x^2 + 8x = 2x^2 + 6x + 4$. Canceling and combining like terms yields $x=2$. 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 $a/(b+c), b/(c+a), c(a+b)$ is 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 problems that utilize harmonic sequences and series.

Introductory

See Also

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