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Difference between revisions of "Harmonic sequence"

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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]].
 
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]].
  
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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.
+
For example, <math>-1, -\frac{1}{2}, -\frac{1}{3}, -\frac{1}{4}</math> and <math>6, 3, 2, \frac{6}{4}</math> are harmonic sequences; however, <math>0, \frac{1}{3}, \frac{1}{6}, \frac{1}{9}</math> and <math>\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots</math> are not.
  
 
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>.
 
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>.
  
 
== Properties ==
 
== 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>
+
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 finite 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>
  
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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.
+
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 ==
 
== 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.
+
A '''harmonic series''' is the sum of all the terms in a harmonic series. All infinite harmonic series diverges, which follows by the limit comparison test with the series <math>1 + 1/2 + 1/3 + \cdots</math>. This series is referred to as ''the'' [[harmonic series]]. As for finite harmonic series, there is no known general expression for their sum; one must find a strategy to evaluate one on a case-by-case basis.
  
 
== Examples ==
 
== Examples ==
−
Here are some example solutions that utilize harmonic sequences and series.
+
Here are some example problems that utilize harmonic sequences and series.
  
 
=== Example 1 ===
 
=== Example 1 ===
 
''Find all real numbers such that <math>x+4, x+1, x</math> is a harmonic sequence.''
 
''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>
+
'''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>, which simplifies to <math>x=2</math>. Thus, <math>2, 3, 6</math> is the only solution to the equation, as desired. <math>\square</math>
  
 
=== Example 2 ===
 
=== 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.''
+
''Let <math>a</math>, <math>b</math>, and <math>c</math> be positive real numbers. Show that if <math>a</math>, <math>b</math>, and <math>c</math> are in harmonic progression, then <math>a/(b+c)</math>, <math>b/(c+a)</math>, and <math>c/(a+b)</math> are 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.
 
'''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.
+
One approach might be to add <math>2</math> to both sides of the equation, which when combined with the fractions returns <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> are all 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>
+
From here, it is easy to write the proof forwards. Doing so proves that <math>(b+c)/a + (a+b)/c = 2(c+a)/b</math>, which implies that <math>a/(b+c)</math>, <math>b/(c+a)</math>, <math>c/(a+b)</math> is a harmonic sequence, as required. <math>\square</math>
  
 
=== Example 3 ===
 
=== 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>?
 
''[[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.
+
'''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>. This is the harmonic mean, which implies that <math>a_{n-1}, a_n, a_{n+1}</math> is a harmonic progression. Thus, the entire sequence is in harmonic progression.
  
−
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>
+
Using the tools of harmonic sequences, we will now find a [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT] 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>. <math>\square</math>
  
−
=== More Problems ===
+
== More Problems ==
−
Here are some problems that utilize harmonic sequences and series.
+
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 ===
 
=== Introductory ===
Line 44: Line 44:
  
 
== See Also ==
 
== See Also ==
−
* [[Harmonic series]]
 
 
* [[Arithmetic sequence]]
 
* [[Arithmetic sequence]]
 
* [[Geometric sequence]]
 
* [[Geometric sequence]]
 +
* [[Harmonic series]]
 
* [[Sequence]]  
 
* [[Sequence]]  
 
* [[Series]]
 
* [[Series]]
  
 
[[Category:Algebra]] [[Category:Sequences and series]] [[Category:Definition]]
 
[[Category:Algebra]] [[Category:Sequences and series]] [[Category:Definition]]

Latest revision as of 15:50, 19 February 2024

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 $6, 3, 2, \frac{6}{4}$ are harmonic sequences; however, $0, \frac{1}{3}, \frac{1}{6}, \frac{1}{9}$ and $\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots$ are not.

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 finite 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, which follows by the limit comparison test with the series $1 + 1/2 + 1/3 + \cdots$. This series is referred to as the harmonic series. As for finite harmonic series, there is no known general expression for their sum; one must find a strategy to evaluate one on a case-by-case basis.

Examples

Here are some example problems 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 when combined with the fractions returns \[\frac{a+b+c}{a} + \frac{a+b+c}{c} = \frac{2(a+b+c)}{b}.\] Because $a$, $b$, and $c$ are all 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. Doing so proves that $(b+c)/a + (a+b)/c = 2(c+a)/b$, which implies that $a/(b+c)$, $b/(c+a)$, $c/(a+b)$ is a harmonic sequence, 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, $2/a_{n-1} = 1/a_{n-2} + 1/a_n$. This is the harmonic mean, which implies that $a_{n-1}, a_n, a_{n+1}$ is a harmonic progression. Thus, the entire sequence is in harmonic progression.

Using the tools of harmonic sequences, we will now find a TOTO SLOT 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 $8075 + 3 = 8078$. $\square$

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

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