The Geometry of Harmonic Means

Harmonic progressions and means appear regularly in HSC Extension 1 sequences and series problems. In this post we’ll translate a harmonic progression into an arithmetic progression using reciprocals, derive the formula for the harmonic mean, prove the classic ordering (H<G<A) for two distinct positive numbers, and finally demonstrate why the harmonic series diverges even though its terms shrink to zero. Working through these steps will strengthen your algebraic manipulation and give you a reliable toolkit for tougher sequences questions.

Problem Statement

A sequence $(h_n)$ is in harmonic progression (HP) if the sequence of reciprocals

$$x_n=\frac{1}{h_n}$$

forms an arithmetic progression (AP).

• Given that the third term of an HP is $1$ and the sixth term is $\frac12$, find the first term $h_1$ and the common difference $d$ of the underlying AP.
• The harmonic mean $H$ of two positive numbers $a$ and $b$ is defined so that $a,H,b$ is an HP. Show algebraically that

$$H=\frac{2ab}{a+b}.$$

• Let

$$A=\frac{a+b}{2},\qquad G=\sqrt{ab}$$

be the arithmetic and geometric means of $a$ and $b$. Prove that $G^2=AH$. Hence prove that, for distinct positive real numbers $a$ and $b$,

$$H<G<A.$$

• Consider the infinite harmonic series

$$\sum_{n=1}^{\infty}\frac1n.$$

By comparing the sum of the first $2^k$ terms to blocks of constant lower bounds, prove that this harmonic series does not have a finite limiting sum.

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Hints

• For (i): Convert the HP terms into AP terms $x_3$ and $x_6$, then use $x_n=a+(n-1)d$.
• For (ii): If $a,H,b$ is an HP, then $\frac1a,\frac1H,\frac1b$ is an AP.
• For (iii): Multiply $A$ and $H$. For the inequality, use $(a-b)^2>0$ or $(\sqrt a-\sqrt b)^2>0$.
• For (iv): Group terms as $\left(\frac13+\frac14\right)$, then $\left(\frac15+\cdots+\frac18\right)$, and so on.

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Solutions

(i) We first rewrite the given HP terms using their reciprocals, so that the $x_n$ form an arithmetic progression. Since $x_n=\frac1{h_n}$,

$$x_3=1,\qquad x_6=2.$$

For the underlying AP, the difference $x_6-x_3$ must be $3d$, giving

$$x_6-x_3=3d \implies 2-1=3d \implies d=\frac13.$$

Now we can step backwards from $x_3$ to find $x_1$:

$$x_1=x_3-2d=1-\frac23=\frac13,$$

so

$$h_1=\frac1{x_1}=3.$$

(ii) By the definition of a harmonic progression, if $a,H,b$ is an HP, then $\frac1a,\frac1H,\frac1b$ is an AP. In any arithmetic progression the middle term is the average of the two outer terms, so we have

$$\frac1H=\frac12\left(\frac1a+\frac1b\right)=\frac{a+b}{2ab}.$$

Taking reciprocals of both sides yields the harmonic mean formula:

$$H=\frac{2ab}{a+b}.$$

(iii) We start by examining the product of the arithmetic and harmonic means. Multiplying their formulas directly,

$$AH=\frac{a+b}{2}\cdot\frac{2ab}{a+b}=ab=G^2.$$

Thus $G^2=AH$. For the inequality, we exploit the fact that for distinct positive numbers $a$ and $b$ the square of their difference is positive. In particular,

$$(\sqrt a-\sqrt b)^2>0 \implies a+b>2\sqrt{ab} \implies A>G.$$

Since $G^2=AH$ and all quantities are positive, $A>G$ implies

$$H=\frac{G^2}{A}<G.$$

Combining these gives the full ordering

$$H<G<A.$$

(iv) The harmonic series $\sum\frac1n$ is a classic example of a divergent series whose terms tend to zero. To prove divergence we group the terms into dyadic blocks, each of which can be shown to exceed $\frac12$ after the first two terms. Starting with the first two terms individually, we write

$$1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\cdots+\frac18\right)+\cdots.$$

Now estimate each block from below by replacing every term with the smallest term in that block:

$$\frac13+\frac14>\frac14+\frac14=\frac12,\qquad \frac15+\cdots+\frac18>4\cdot\frac18=\frac12,$$

and in general

$$\frac1{2^{j-1}+1}+\cdots+\frac1{2^j}>2^{j-1}\cdot\frac1{2^j}=\frac12.$$

Therefore the partial sums exceed

$$1+\frac12+\underbrace{\frac12+\frac12+\cdots+\frac12}_{\text{arbitrarily many terms}},$$

so the sequence of partial sums grows without bound. Hence the harmonic series has no finite limiting sum.

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Takeaways

• Harmonic progressions are arithmetic progressions viewed through reciprocals.
• For positive distinct numbers, the classical means satisfy $H<G<A$.
• A sequence can have terms tending to zero while its infinite series still diverges.

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Further Readings

HSC Trigonometry, HSC Last Resorts, HSC Inequalities, HSC Polys Ext 1