Difference between revisions of "Harmonic sequence"
Etmetalakret (talk | contribs) |
Etmetalakret (talk | contribs) m (Corrected minor errors in solution 2) |
||
Line 26: | Line 26: | ||
'''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 | + | One approach might be to add <math>2</math> to both sides of the equation, which gives us <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. | + | From here, it is easy to write the proof forwards. Doing so yields 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>, and <math>c/(a+b)</math> is in harmonic progression, as required. <math>\square</math> |
=== Example 3 === | === Example 3 === |
Revision as of 21:45, 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. By definition,
can never be a term of a harmonic sequence.
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
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 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; this is by a limit comparison test with the series , 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 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
, or
.
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.