The Geometry of Harmonic Means

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}.

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.

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.

Solutions

(i) Since $x_n=\frac1{h_n}$ ,

x_3=1,\qquad x_6=2.

For the underlying AP,

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

Also

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

h_1=\frac1{x_1}=3.

(ii) Since $\frac1a,\frac1H,\frac1b$ is an AP, the middle term is the average of the outer terms:

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

Therefore

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

(iii) Using the formulas for $A$ and $H$ ,

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

For $a\ne b$ ,

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

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

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

Hence

H<G<A.

(iv) Group the harmonic series into dyadic blocks:

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

Each block after the first has sum greater than or equal to $\frac12$ :

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

and similarly

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

Thus the partial sums exceed

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

so they grow without bound. Therefore the harmonic series has no finite limiting sum.

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.

Further Readings

HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/
HSC Last Resorts: https://vumaths.com/booklets/hsc-last-resorts/
HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/
HSC Polys Ext 1: https://vumaths.com/booklets/hsc-polys-ext-1/

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