Difference between revisions of "Harmonic sequence"
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'''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>. 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 | + | 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>. <math>\square</math> |
== More Problems == | == More Problems == |
Revision as of 21:13, 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, and are harmonic sequences; however, and are not.
More formally, a harmonic progression biconditionally satisfies A similar definition holds for infinite harmonic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in harmonic progression if and only if .
Contents
[hide]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 and , the terms of any finite harmonic sequence can be written as
A common lemma is that a sequence is in harmonic progression if and only if is the harmonic mean of and for any consecutive terms . In symbols, . 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 . 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 it 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 is a harmonic sequence.
Solution: Using the harmonic mean properties of harmonic sequences, Note that would create a term of —something that breaks the definition of harmonic sequences—which eliminates them as possible solutions. We can thus multiply both sides by to get . Expanding these factors yields , which simplifies to . Thus, is the only solution to the equation, as desired.
Example 2
Let , , and be positive real numbers. Show that if , , and are in harmonic progression, then , , and are as well.
Solution: Using the harmonic mean property of harmonic sequences, we are given that , and we wish to show that . We work backwards from the latter equation.
One approach might be to add to both sides of the equation, which gives us Because , , and were all defined to be positive, . Thus, we can divide both sides of the equation by to get , which was given as true.
From here, it is easy to write the proof forwards. Doing so yields that , which implies that , , and is in harmonic progression, as required.
Example 3
2019 AMC 10A Problem 15: A sequence of numbers is defined recursively by , , and for all Then can be written as , where and are relatively prime positive integers. What is ?
Solution: We simplify the series recursive formula. Taking the reciprocals of both sides, we get the equality Thus, . 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 and . Simplifying the first equation yields and substituting this into the second equation yields . Thus, and so . The answer is then .
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.