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,

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

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 $\displaystyle\sum_{i=1}^{n} \frac{d}{a+(i-1)d}:$

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

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

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

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

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

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