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 \(1,2,3,4,5,6, \ldots\) is an arithmetic progression, so its reciprocals \(\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\) are a 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. \(_\square\)

A sequence is a harmonic progression if and only if its terms are the reciprocals of an arithmetic progression that doesn't contain \( 0.\) \(_\square\)

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. \(_\square\)

If the \(3^\text{rd}\) term of an HP is \(3\) and the \(10^\text{th}\) term of the same HP is \(10\), what is the maximum possible length of the HP?

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The reciprocals of the HP form an arithmetic progression \( a,a+d,a+2d,\ldots\). Then \(a + 2d = \frac{1}{3}\) and \(a + 9d = \frac{1}{10}.\)

This gives \(a =\frac{2}{5}\) , \(d = -\frac{1}{30}.\) Note that \( a+12d = 0,\) so the arithmetic progression can only have \( 12\) terms before it contains \( 0.\) So the answer is \( 12.\) \(_\square\)

\((\)Exercise: If the \(m^\text{th}\) term is \(m\) and the \(n^\text{th}\) term is \(n,\) the answer is \( m+n-1.)\)

If the first \(2\) terms of a harmonic progression \( a_n\) are \( \frac{1}{19}\) and \(\frac{1}{17}, \) find the maximum partial sum \( \sum\limits_{n=1}^k a_n. \)

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The terms of the HP are

\[\frac{1}{19},\ \frac{1}{17},\ \frac{1}{15},\ \ldots,\ \frac{1}{3},\ \frac{1}{1},\ \frac{1}{-1},\ \frac{1}{-3},\ \ldots.\]

So the maximum partial sum is

\[\frac{1}{19} + \frac{1}{17} + \frac{1}{15} +\cdots + \frac{1}{1} \approx 2.13.\ _\square\]

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 \(\frac{17}{70}\), the sum of the next \(2\) terms is \(\frac{5}{4}\), and the sum of the following \(2\) terms is \(\frac{-7}{10}\), find the sum of the following \(2\) terms (again).

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According to the problem statement,

\[\begin{align} \frac{1}{a} + \frac{1}{(a+d)} &= \frac{17}{70} &\qquad (1)\\ \frac{1}{(a+2d)} + \frac{1}{(a+3d)} &= \frac{5}{4} &\qquad (2)\\ \frac{1}{(a+4d)} + \frac{1}{(a+5d)} &= \frac{-7}{10}. &\qquad (3) \end{align}\]

Solving any two of \((1), (2),\) and \((3)\) simultaneously gives \(a = 10\) and \(d = -3\).

Therefore, the answer is

\[\frac{1}{(a+6d)} + \frac{1}{(a+7d)} = \frac{-19}{88}.\ _\square\]

The partial sums of a harmonic progression are often of interest. For example, the harmonic numbers are partial sums of the harmonic progression \( 1, \frac12, \frac13, \ldots\). 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 \(\frac1{a},\frac1{a+d},\ldots,\frac1{a+(n-1)d},\) where \( d>0\).

Consider the function \(f(x)=\frac{1}{x}\), and consider Riemann sums with rectangles of width \(d\) starting from \(x=a\) to \(x=a+(n-1)d\).

Here, the \(i^\text{th}\) rectangle has length \(\frac1{a+(i-1)d}\), and the width of each rectangle is a constant \(d\). So the sum of the areas 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 \(x=a+(n-\frac{1}{2})d\) is approximately \(\displaystyle\sum_{i=1}^{n} \frac{d}{a+(i-1)d}:\)

\[\begin{align} \int_{a-\frac{1}{2}}^{a+\big(n-\frac{1}{2}\big)d} \frac1{x}\,\mathrm{d}x &≈ \sum_{i=1}^{n} \frac{d}{a+(i-1)d}\\ \int_{a-\frac{1}{2}}^{a+\big(n-\frac{1}{2}\big)d} \frac{1}{x}\,\mathrm{d}x &≈ d \sum_{i=1}^{n} \frac1{a+(i-1)d}\\ d S_n &≈ \int_{a-\frac{1}{2}}^{a+\big(n-\frac{1}{2}\big)d} \frac{1}{x}\,\mathrm{d}x\\ S_n &≈ \left. \frac1{d} \ln(x) \right |_{a-\frac{1}{2}}^{a+\big(n-\frac{1}{2}\big)d}\\ S_n &≈ \frac1{d} \ln \left(\frac{2a+(2n-1)d}{2a-d} \right), \end{align}\]

where \(2a ≠ d\) 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 \( \left(\frac{1}{19} + \frac{1}{17} + \cdots + \frac{1}{3}\right) + 1.\) Apply the formula to the parenthetical sum, and rearrange terms so that \(d\) is positive, i.e. \( a=3, d=2, n=9.\)

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The formula gives

\[\begin{align} S_n &≈ 1+ \frac{1}{2} \ln \left(\frac{2(3)+(2(9)-1)(2)}{2(3)-(2)} \right)\\ &≈ 1+ \frac{1}{2} \ln \left(\frac{40}{4} \right)\\ &≈ 1+\frac{1}{2} \ln(10)\\ &≈ 2.15. \end{align}\]

This is indeed a reasonable approximation to the exact value given above. \(_\square\)

Another nice approximation of \(S_n\) can be obtained in the following way:

Taking \(b=\frac{a}{d}\), note that

\[\begin{align} \lim_{n\to \infty}\left(S_n-\frac{1}{d}\ln\left(\frac{a+dn}{a+d}\right)\right) &=\frac{1}{d}\lim_{n\to \infty}\left[\sum_{k=1}^n\frac{1}{b+k}-\ln\left(\frac{b+n}{b+1}\right)\right]\\ &=\frac{1}{d}\lim_{n\to \infty}\left[\sum_{k=1}^n\left(\frac{1}{b+k}-\frac{1}{k}\right)+\ln n-\ln\left(\frac{b+n}{b+1}\right)+\sum_{k=1}^n\frac{1}{k}-\ln n\right]\\ &=-\frac{1}{d}\lim_{n\to \infty}\left(\sum_{k=1}^n \frac{1}{k}-\frac{1}{b+k}\right)+\frac{\ln(b+1)}{d}+\frac{1}{d}\lim_{n\to \infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln n\right). \end{align}\]

The first term can be obtained from the digamma function as

\[\psi(s+1)=-\gamma+\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{s+k}\right),\]

where \(\gamma\) is the Euler-Mascheroni constant, which is defined as

\[\lim_{n\to \infty}(H_n-\ln n)=\gamma\approx 0.57722,\]

where \(H_n\) is the harmonic number, and thus simplifies the third term. So, we get

\[\lim_{n\to \infty}\left(S_n-\frac{1}{d}\ln\left(\frac{a+dn}{a+d}\right)\right)=\frac{1}{d}\big[-\gamma-\psi(b+1)+\ln(b+1)+\gamma\big]=\frac{\ln\left(1+\frac{a}{d}\right)-\psi\left(1+\frac{a}{d}\right)}{d}.\]

Thus, for large \(n\), we get the approximation \(S_n\approx \frac{1}{d}\left(\ln\left(n+\frac{a}{d}\right)-\psi(1+\frac{a}{d})\right)\).

• Arithmetic Progression

• Arithmetic-geometric Progression

• Geometric Progression

• Harmonic Series

• Power Mean Inequality (QAGH)

• Harmonic Number