Harmonic Progression
A harmonic progression is a sequence of real numbers formed by taking the reciprocals of an arithmetic progression. Equivalently, it is a sequence of real numbers such that any term in the sequence is the harmonic mean of its two neighbors.
The sequence \(1,2,3,4,....\) is an arithmetic progression, so its reciprocals \(\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},...\) are a harmonic progression.
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Formal definition of harmonic progression
A sequence \( a_n\) of real numbers is a harmonic progression (HP) if any term in the sequence is the harmonic mean of its two neighbors.
A sequence is a harmonic progression if and only if its terms are the reciprocals of an arithmetic progression that doesn't contain \( 0.\)
The harmonic mean of \( \frac1{a+(n-1)d}\) and \( \frac1{a+(n+1)d} \) is \[ \frac2{a+(n-1)d+a+(n+1)d} =\frac2{2a+2nd}= \frac1{a+nd}, \] which proves the "if" part of the statement. The "only if" part is similar and left as an exercise: write the first two terms as \( \frac1{a}\) and \(\frac1{a+d},\) and show that the third term is \( \frac1{a+2d},\) the fourth term is \( \frac1{a+3d},\) and so on by induction.
Problem-solving and examples with harmonic progressions
If the \(3^{rd}\) term of a HP is \(3\) and the \(10^{th}\) term of the same HP is \(10\), what is the maximum possible length of the HP?
The reciprocals of the HP form an arithmetic progression \( a,a+d,a+2d,\ldots\) Then \(a + 2d = 1/3\) and \(a + 9d = 1/10.\)
This gives \(a =2/5\) , \(d = -1/30.\) Note that \( a+12d = 0,\) so the arithmetic progression can only have \( 12\) terms before it contains \( 0.\) So the answer is \( 12.\)
(Exercise: if the \(m\)th term is \(m\) and the \(n\)th term is \(n,\) the answer is \( m+n-1.)\)
If the first \(2\) terms of a harmonic progression \( a_n\) are \( \dfrac{1}{19}, \dfrac{1}{17}, \) find the maximum partial sum \( \sum\limits_{n=1}^k a_n. \)
The terms of the HP are,
\(1/19, 1/17, 1/15, \ldots, 1/3,1/1, 1/-1, 1/-3, \ldots\)
So the maximum partial sum is \[ 1/19 + 1/17 + 1/15 + ... + 1/1 \approx 2.13. \]
The next section will give a general approximation for partial sums of harmonic progressions.
If the sum of the first \(2\) terms of an HP is \(17/70\) and the sum of the next \(2\) terms is \(5/4\), and the sum of the following \(2\) terms is \(-7/10\), find the sum of the following \(2\) terms (again).
According to the question,
\(1/a + 1/(a+d) = 17/70 \text{...} (1)\)
\(1/(a+2d) + 1/(a+3d) = 5/4 \text{...} (2)\)
\(1/(a+4d) + 1/(a+5d) = -7/10 \text{...} (3)\)
Solving any two of \((1)\),\((2)\) and \((3)\) simultaneously gives \(a = 10\) and \(d = -3\).
The answer is \(1/(a+6d) + 1/(a+7d) = \boxed{-19/88}.\)
Sum of terms of harmonic progression (Approximation)
The partial sums of a harmonic progression are often of interest. For example, the harmonic numbers are partial sums of the harmonic progression \( 1, 1/2, 1/3, \ldots.\) The following approximation for the partial sums of a harmonic progression is due to Brillianteer Aneesh Kundu.
The goal is to find the sum \( S_n \) of the terms \(\frac1{a},\frac1{a+d},...\frac1{a+(n-1)d}.\) Assume that \( d\) is positive.
Consider the function \(f(x)=1/x\), and consider Riemann sums with rectangles of width \(d\) starting from \(x=a\) to \(x=a+(n-1)d\).
Here the \(i^{th}\) rectangle has length \(\frac1{a+(i-1)d}\) and the width of each rectangle is a constant \(d\). So the sum of area of these rectangles will be approximately equal to the area under the curve.
That is, the area under the curve from \(x=a-\frac{d}{2}\) to \(x=a+(n-\frac{1}{2})d\) \(≈ \displaystyle\sum_{i=1}^{n} \frac{d}{a+(i-1)d}\)
\(\displaystyle\int_{a-\frac{1}{2}}^{a+(n-\frac{1}{2})d} \frac1{x}\,\mathrm{d}x≈ \displaystyle\sum_{i=1}^{n} \frac{d}{a+(i-1)d}\)
\(\displaystyle\int_{a-\frac{1}{2}}^{a+(n-\frac{1}{2})d} \frac{1}{x}\,\mathrm{d}x≈ d\displaystyle\sum_{i=1}^{n} \frac1{a+(i-1)d}\)
\(d S_n ≈ \displaystyle\int_{a-\frac{1}{2}}^{a+(n-\frac{1}{2})d} \frac{1}{x}\,\mathrm{d}x\)
\(S_n ≈ \left. \frac1{d} \ln(x) \right |_{a-\frac{1}{2}}^{a+(n-\frac{1}{2})d}\)
\(S_n ≈ \frac1{d} \ln \left(\frac{2a+(2n-1)d}{2a-d} \right)\)
Here,
\(2a ≠ d\) and \(d ≠ 0\).
This formula gives a good approximation when \(n\) is very big as compared to \(d.\)
The example from the previous section involved computing \( (1/19 + 1/17 + \cdots + 1/3) + 1.\) Apply the formula to the parenthetical sum, and rearrange terms so that \(d\) is positive, i.e. \( a=3, d=2, n=9.\)
The formula gives
\(S_n ≈ 1+ \frac{1}{2} \ln \left(\frac{2(3)+(2(9)-1)(2)}{2(3)-(2)} \right)\)
\(S_n ≈ 1+ \frac{1}{2} \ln \left(\frac{40}{4} \right)\)
\(S_n ≈ 1+\frac{1}{2} \ln(10)\)
\(S_n ≈ 2.15\)
This is indeed a reasonable approximation to the exact value given above.