Difference between revisions of "Harmonic sequence"

Revision as of 14:33, 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 examples 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 $E$ . $\square$

@@ Line 1: / Line 1: @@
 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, <math>1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}</math> and <math>\frac{1}{99}, \frac{1}{91}, \frac{1}{83}, \frac{1}{75}</math> are harmonic sequences; however, <math>1, 1, \frac{1}{3}, \frac{1}{5}</math> and <math>\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots</math> are not.
+For example, <math>1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}</math> and <math>\frac{1}{99}, \textrm{ } \frac{1}{91}, \textrm{ } \frac{1}{83}, \frac{1}{75}</math> are harmonic sequences; however, <math>1, 1, \frac{1}{3}, \frac{1}{5}</math> and <math>\frac{1}{4}, \frac{1}{12},  \frac{1}{36}, \frac{1}{108}, \ldots</math> are not. By definition, <math>0</math> can never be a term of a harmonic sequence.
-More formally, the sequence <math>a_1, a_2, \ldots , a_n</math> is a harmonic progression if and only if <math>\frac{1}{a_2} - \frac{1}{a_1} = \frac{1}{a_3} - \frac{1}{a_2} = \cdots = \frac{1}{a_n} - \frac{1}{a_{n-1}}.</math> A similar definition holds for infinite harmonic sequences. It appears most frequently in its three-term form: namely, that constants <math>a</math>, <math>b</math>, and <math>c</math> are in harmonic progression if and only if <math>\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}</math>.
+More formally, a harmonic progression <math>a_1, a_2, \ldots , a_n</math> biconditionally satisfies <math>1/a_2 - 1/a_1 = 1/a_3 - 1/a_2 = \cdots = 1/a_n - 1/a_{n-1}.</math> A similar definition holds for infinite harmonic sequences. It appears most frequently in its three-term form: namely, that constants <math>a</math>, <math>b</math>, and <math>c</math> are in harmonic progression if and only if <math>1/b - 1/a = 1/c - 1/b</math>.
-WIP
+== 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 <math>a</math> and <math>d</math>, the terms of any harmonic sequence can be written as <cmath>\frac{1}{a}, \textrm{ } \frac{1}{a+d}, \textrm{ } \frac{1}{a+2d}, \textrm{ } \cdots \textrm{ } \frac{1}{a+(n-1)d}.</cmath>
+A common lemma is that a sequence is in harmonic progression if and only if <math>a_n</math> is the harmonic mean of <math>a_{n-1}</math> and <math>a_{n+1}</math>  for any consecutive terms <math>a_{n-1}, a_n, a_{n+1}</math>. In symbols, <math>2/a_n = 1/a_{n-1} + 1/a_{n+1}</math>. 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 <math>1 + 1/2 + 1/3 + \cdots</math>, 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 examples that utilize harmonic sequences and series.
+=== Example 1 ===
+''Find all real numbers such that <math>x+4, x+1, x</math> is a harmonic sequence.''
+'''Solution''': Using the harmonic mean properties of harmonic sequences, <cmath>\frac{2}{x+1} = \frac{1}{x+4} + \frac{1}{x} = \frac{2x+4}{x(x+4)}.</cmath> Note that <math>x=-4, -1, 0</math> would create a term of <math>0</math>—something that breaks the definition of harmonic sequences—which eliminates them as possible solutions. We can thus multiply both sides by <math>x(x+1)(x+4)</math> to get <math>2x(x+4) = (2x+4)(x+1)</math>. Expanding these factors yields <math>2x^2 + 8x = 2x^2 + 6x + 4</math>. Canceling and combining like terms yields <math>x=2</math>. Thus, <math>2, 3, 6</math> is the only solution to the equation, as desired. <math>\square</math>
+=== Example 2 ===
+''Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers. Show that if <math>a, b, and c</math> are in harmonic progression, then <math>a/(b+c), b/(c+a), c(a+b)</math> is as well.''
+'''Solution''': Using the harmonic mean property of harmonic sequences, we are given that <math>1/a + 1/b = 2/c</math>, and we wish to show that <math>(b+c)/a + (a+b)/c = 2(c+a)/b</math>. We work backwards from the latter equation.
+One approach might be to add <math>2</math> to both sides of the equation, which yields <cmath>\frac{a+b+c}{a} + \frac{a+b+c}{c} = \frac{2(a+b+c)}{b}.</cmath> Because <math>a</math>, <math>b</math>, and <math>c</math> were all defined to be positive, <math>a+b+c \neq 0</math>. Thus, we can divide both sides of the equation by <math>a+b+c</math> to get <math>1/a + 1/c = 2/b</math>, which was given as true.
+From here, it is easy to write the proof forwards. Then <math>(b+c)/a + (a+b)/c = 2(c+a)/b</math>, which implies that the sequence <math>a/(b+c), b/(c+a), c(a+b)</math> is in harmonic progression, as required. <math>\square</math>
+=== Example 3 ===
+''[[2019 AMC 10A Problems/Problem 15 | 2019 AMC 10A Problem 15]]: A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and <math>a_n = \frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2} - a_{n-1}}</math> for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q</math>?
+'''Solution''': We simplify the series recursive formula. Taking the reciprocals of both sides, we get the equality <cmath>\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}}.</cmath> Thus, <math>2/a_{n_1} = 1/a_{n-2} + 1/a_n</math>. 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 <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>
+== See Also==
+* [[Harmonic series]]
+* [[Arithmetic sequence]]
+* [[Geometric sequence]]
+* [[Sequence]]
+* [[Series]]
+[[Category:Algebra]] [[Category:Sequences and series]] [[Category:Definition]]

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