# Harmonic Sequence and Series

In mathematics, a harmonic progression (HP or harmonic sequence HS) is a progression formed by taking the reciprocals of an arithmetic progression.
The sequence 1,2,3,4,5,6,... is an arithmetic progression, so its reciprocals
$\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$ are a harmonic progression.

If each term of an harmonic sequence is multiplied or divide by a constant, the sequence of the resulting number are also harmonic sequence. For example,

1. if a,b,c,d,... are harmonic sequence
2. then ka,kb,kc,kd,... are harmonic sequence where k is constant and $k \ne 0$
3. $\frac{a}{k},\frac{b}{k},\frac{c}{k},\frac{d}{k},...$ are harmonic sequence where $k \ne 0$

#### General term of Harmonic sequence

Let a,a+d,a+2d,a+3d, .... are the terms of arithemetic sequence
Then
terms of Harmonic sequence are given by
$\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}+\frac{1}{a+3d},...$
Now,
The nth term of the sequence is given by
$a_n=\frac{1}{a+(n-1)d}$

#### Harmonic Mean

Harmonic mean is finding as to the reciprocal of the arithmetic mean of the reciprocals. The formula to calculate the harmonic mean is given by:
Harmonic Mean =$\frac{n}{(\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})+... }$
Where a,b,c,d are the values, and n is the number of values present.

#### Sum of Harmonic Series

If $\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d}$ is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, $S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]$
where,
“a” is the first term of A.P
“d” is the common difference of A.P
“ln” is the natural logarithm

Proof
Given $\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d}$ are in harmonic progression,
We set the values as follows,
a=1,d=1
Then, $\frac{1}{1},\frac{1}{1+1},\frac{1}{1+2.1},…,\frac{1}{1+(n-1)1}$ are the terms
$1,\frac{1}{2},\frac{1}{3}, ...,$are the terms
Now, the Riemann sum of the function $f(x)=\frac{1}{x}$ approximates the sum of the harmonic series given above
Therefore,
We set the values as follows,
$S_n= \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx$
For any common difference,the formula becoms
$S_n=\frac{1}{d} \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx$
or $S_n= \frac{1}{d} \int_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx$
or $S_n= \frac{1}{d} \left [ \ln x \right]_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d}$
or $S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]$
Therefore The formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, $S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]$

#### A.M, G.M, H.M and their relations

A finite sequence consisting ore than two terms has one or more terms in between the first and last terms. Theses between terms are called means of the sequence. Precisely

1. if a,b,c,d are in arithmetic sequence then b,c are arithmetic means (AM)
2. item if a,b,c,d are in geometric sequence then b,c are geometric means (GM)
3. if a,b,c,d are in harmonic sequence then b,c are harmonic means (HM)

#### Theorem 1

Given any two numbers a and b the AM, GM, and HM are as follows.

1. AM=$\frac{a+b}{2}$
2. GM=$\sqrt{ab}$
3. HM=$\frac{2ab}{a+b}$

The proof are as follows:

1. Let AM is the single mean between a and b, then
AM-a=b-AM
or 2AM=a+b
or AM=$\frac{a+b}{2}$
2. Let GM is the single mean between a and b, then
$\frac{GM}{a}=\frac{b}{GM}$
or $GM^2=ab$
or $GM=\sqrt{ab}$
3. Let HM is the single mean between a and b, then
$\frac{1}{HM}-\frac{1}{a}=\frac{1}{b}-\frac{1}{HM}$
or $\frac{2}{HM}=\frac{1}{a}+\frac{1}{b}$
or $HM=\frac{2ab}{a+b}$

#### Theorem 2

Let a and b are two non-negative numbers, then

1. $GM^2=AM \times HM$
2. $AM \ge GM \ge HM$ Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.

Let a and b are two non-negative numbers then,
$AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}$
Now, we have
$GM^2=ab$
or $GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b}$
or $GM^2=AM \times HM$

For the second part, we write
$AM-GM=\frac{a+b}{2}-\sqrt{ab}$
or $AM-GM=\frac{a+b-2\sqrt{ab}}{2}$
or $AM-GM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}-2\sqrt{a}\sqrt{b}}{2}$
or $AM-GM=\frac{{{( \sqrt{a}-\sqrt{b} )}^{2}}}{2}$
or $AM\ge GM$ (1)
Similarly,
or $GM-HM=\sqrt{ab}-\frac{2ab}{a+b}$
or $GM-HM=\frac{\sqrt{ab}( a+b )-2ab}{a+b}$
or $GM-HM=\frac{\sqrt{ab}( a+b )-2\sqrt{ab}\sqrt{ab}}{a+b}$
or $GM-HM=\frac{\sqrt{ab}}{a+b}( a+b-2\sqrt{ab} )$
or $GM-HM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}-\sqrt{b} )}^{2}}$
or $GM\ge HM$ (2)
Combining (1) and (2), we get
or $AM\ge GM\ge HM$

#### Visualization of the proof

Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.
Then OP is the arithemetic mean given by
$AM =\frac{a+b}{2}$

#### Visualization of AM,GM and HM

By the property of radius and diameter, we get that
$AM =\frac{a+b}{2}$

Also, by the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by
$GM =\sqrt{ab}$

By using proportionality, we get
Triangle ADQ and QDB are similar with AD=a, DB=b, so we have
$\frac{GM}{a}=\frac{b}{GM}$
or $GM= \sqrt{ab}$

Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by
$HM =\frac{2ab}{a+b}$

By using proportionality, we get
Triangle DRQ and ODQ are similar with QR=GM,QD=$\sqrt{ab}$, OD=$\frac{a-b}{2}$, so we have
$\frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}}$
or $HM= \frac{2ab}{a+b}$