Irrationality of π(PI): Lambert's Proof

Introduction

After mentioning about the Lambert's famous proof of irrationality of $\pi$ in an earlier post, it is now time to give it to the readers in its entirety. I need to reiterate the fact that being a far more direct proof than the modern proofs of Ivan Niven, it is still highly neglected by modern authors and educators. The idea of the proof is really elementary but based on the concept of continued fractions which are now deleted from the high school mathematical syllabus. Why this topic is now left out is still unclear to me. One reason which I can guess of is that the manipulations of continued fractions are not so simple (compared to those of an infinite product or a series). The visible form of the continued fraction does not give any idea about its value unless we do the calculations.

Value of an Infinite Continued Fraction

We shall first introduce some standard notation for continued fractions and other quantities related with it. To start with we will write the continued fraction in the form on the left of the equation below and it should be taken as representing the form given on the right. This way of writing saves vertical space.
$$a_{1} + \dfrac{b_{2}}{a_{2} +}\,\dfrac{b_{3}}{a_{3} +}\,\dfrac{b_{4}}{a_{4} +}\,\dfrac{b_{5}}{a_{5} +}\,\cdots = a_{1} + \dfrac{b_{2}}{a_{2} + \dfrac{b_{3}}{a_{3} + \dfrac{b_{4}}{a_{4} + \dfrac{b_{5}}{a_{5} + \cdots}}}}$$ It is normally preferable to keep all the quantities $b_{i}$ as positive and if they turn out to be negative then we just replace the 'plus' sign before it with a 'minus' sign. The value of a continued fraction is difficult to describe when the continued fraction is infinite (i.e. when the sequences $\{a_{n}\},\, \{b_{n}\}$ are infinite). To that end we define the convergents (its just another sequence of numbers related with the continued fraction in a particular way) $p_{n}/q_{n}$ as follows: \begin{align} p_{0} &= 1, p_{1} = a_{1}, q_{1} = 1, q_{2} = a_{2}\notag\\ p_{n} &= a_{n}p_{n - 1} + b_{n}p_{n - 2},\,\,\,\, n \geq 2 \notag\\ q_{n} &= a_{n}q_{n - 1} + b_{n}q_{n - 2},\,\,\,\, n > 2 \notag \end{align} It is expected that at most only a finite number of $q_{n}$ can vanish, otherwise the sequence of convergents $p_{n} / q_{n}$ has no meaning. Assuming that this is so, we define the value of the infinite continued fraction as the limit of sequence of convergents $p_{n} / q_{n}$ provided this limit exists. In this case we say that the infinite continued fraction converges.

With this brief introduction we can now move on to deal with issues of convergence of specific classes of continued fractions.

Convergence of Continued Fractions

The theory of continued fractions is far less complete when compared with the corresponding theories of infinite series and products. Hence we shall confine ourselves with the following two classes of infinite continued fractions:

Class 1: Infinite continued fractions of the form $$a_{1} + \dfrac{b_{2}}{a_{2} +}\,\dfrac{b_{3}}{a_{3} +}\,\dfrac{b_{4}}{a_{4} +}\,\dfrac{b_{5}}{a_{5} +}\,\cdots$$ where all the numbers $a_{n},\, b_{n}$ are positive.

Class 2: Infinite continued fractions of the form $$a_{1} + \dfrac{b_{2}}{a_{2} -}\,\dfrac{b_{3}}{a_{3} -}\,\dfrac{b_{4}}{a_{4} -}\,\dfrac{b_{5}}{a_{5} -}\,\cdots$$ where all the number $a_{n}\,b_{n}$ are positive.

It is quite clear that we can deduce properties of second class of continued fractions from those of the first class by substituting $-b_{3}, -b_{4}, \cdots$ in place of $b_{3}, b_{4}, \cdots$.

Hence most of the time we will prove properties of first class of continued fractions only. Since we have $p_{n} = a_{n}p_{n - 1} + b_{n}p_{n - 2}$ it follows that $$p_{n} - p_{n - 1} = (a_{n} - 1)p_{n - 1} + b_{n}p_{n - 2}$$ and therefore the fact that:

In an infinite continued fraction of the first class both $p_{n}$ and $q_{n}$ are positive and if $a_{n} \geq 1$ then both the sequences $\{p_{n}\}$ and $\{q_{n}\}$ are strictly increasing sequences.

and similarly one can show that:

In an infinite continued fraction of the second class, if $a_{n} \geq b_{n} + 1$ then $p_{n}$ and $q_{n}$ are positive and both the sequences $\{p_{n}\}$ and $\{q_{n}\}$ are strictly increasing sequences.

We are however more interested in the behavior of the sequence of convergents $p_{n}/q_{n}$ and to that end we notice that \begin{align} p_{n}q_{n - 1} - p_{n - 1}q_{n} &= (a_{n}p_{n - 1} + b_{n}p_{n - 2})q_{n - 1} - p_{n - 1}(a_{n}q_{n - 1} + b_{n}q_{n - 2})\notag\\ &= -b_{n}(p_{n - 1}q_{n - 2} - p_{n - 2}q_{n - 1})\notag \end{align} and from this recursive relation it follows that $$p_{n}q_{n - 1} - p_{n - 1}q_{n} = (-1)^{n}b_{2}b_{3}\cdots b_{n}$$ and therefore $$\frac{p_{n}}{q_{n}} - \frac{p_{n - 1}}{q_{n - 1}} = (-1)^{n}\,\frac{b_{2}b_{3}\cdots b_{n}}{q_{n}q_{n - 1}}$$ Again \begin{align} p_{n}q_{n - 2} - p_{n - 2}q_{n} &= (a_{n}p_{n - 1} + b_{n}p_{n - 2})q_{n - 2} - p_{n - 2}(a_{n}q_{n - 1} + b_{n}q_{n - 2})\notag\\ &= a_{n}(p_{n - 1}q_{n - 2} - p_{n - 2}q_{n - 1})\notag\\ &= (-1)^{n - 1}a_{n}b_{2}b_{3}\cdots b_{n - 1}\text{ (from above)}\notag \end{align} Thus we have the following result: $$\frac{p_{n}}{q_{n}} - \frac{p_{n - 2}}{q_{n - 2}} = (-1)^{n - 1}\,\frac{a_{n}b_{2}b_{3}\cdots b_{n - 1}}{q_{n}q_{n - 2}}$$ It now follows from the above results that
For an infinite continued fraction of the first class, the odd convergents form a strictly increasing sequence whereas the even convergents form a strictly decreasing sequence. Moreover every odd convergent is less than all the convergents following it and every even convergent is greater than all the convergents following it.

In other words we always have the pattern $$p_{2n - 1} / q_{2n - 1} < p_{2n + 1} / q_{2n + 1} < p_{2n} / q_{2n}$$ and $$p_{2n} / q_{2n} > p_{2n + 2} / q_{2n + 2} > p_{2n + 1} / q_{2n + 1}$$ So the sequence of convergents increases and then decreases alternately and each time the difference (in absolute value) between two successive convergents always keeps on decreasing.

It is also clear from the above that both the sequence of odd convergents and even convergents is bounded and therefore they tend to definite limits i.e. $$\lim_{n \to \infty} \frac{p_{2n - 1}}{q_{2n - 1}} = A$$ $$\lim_{n \to \infty} \frac{p_{2n}}{q_{2n}} = B$$ For convergence of the infinite continued fraction it is necessary and sufficient that $A = B$. There is a good chance than this is so because we have already noted that the difference between successive convergents is itself decreasing. We need to ensure that this difference between successive convergents ultimately tends to limit $0$.

To achieve this we need to put further restrictions on the values of $a_{n}, b_{n}$. To that end we need to notice that \begin{align} q_{n} &= a_{n}q_{n - 1} + b_{n}q_{n - 2}\notag\\ q_{n - 1} &= a_{n - 1}q_{n - 2} + b_{n - 1}q_{n - 3} > a_{n - 1}q_{n - 2}\notag\\ q_{n - 2} &= a_{n - 2}q_{n - 3} + b_{n - 2}q_{n - 4} > a_{n - 2}q_{n - 3}\notag\\ \cdots &= \cdots\notag\\ q_{4} &= a_{4}q_{3} + b_{4}q_{2} > a_{4}q_{3}\notag\\ q_{3} &= a_{3}q_{2} + b_{3}q_{1} > a_{3}q_{2}\notag\\ q_{2} &= a_{2}q_{1}\notag \end{align} Hence it follows that \begin{align} q_{n} &> (a_{n}a_{n - 1} + b_{n})q_{n - 2}\notag\\ q_{n - 1} &> (a_{n - 1}a_{n - 2} + b_{n - 1})q_{n - 3}\notag\\ q_{n - 2} &> (a_{n - 2}a_{n - 3} + b_{n - 2})q_{n - 4}\notag\\ \cdots &> \cdots\notag\\ q_{4} &> (a_{4}a_{3} + b_{4})q_{2}\notag\\ q_{3} &= (a_{3}a_{2} + b_{3})q_{1}\notag \end{align} Multiplying these equations we get $$q_{n}q_{n - 1} > q_{1}q_{2}(b_{3} + a_{2}a_{3})(b_{4} + a_{3}a_{4})\cdots (b_{n} + a_{n - 1}a_{n})$$ Noting that $q_{1} = 1, q_{2} = a_{2}$ we get $$\frac{q_{n}q_{n - 1}}{b_{2}b_{3}\cdots b_{n}} = \frac{a_{2}}{b_{2}}\left(1 + \frac{a_{2}a_{3}}{b_{3}}\right)\left(1 + \frac{a_{3}a_{4}}{b_{4}}\right) \cdots \left(1 + \frac{a_{n - 1}a_{n}}{b_{n}}\right)$$ Now suppose that the series $\sum a_{n - 1}a_{n} / b_{n}$ is divergent. Then it follows that the product $\prod (1 + (a_{n - 1}a_{n} / b_{n}))$ is divergent. And hence we have
$$\lim_{n \to \infty}\frac{b_{2}b_{3}\cdots b_{n}}{q_{n}q_{n - 1}} = 0$$ i.e. $$\lim_{n \to \infty} \frac{p_{n}}{q_{n}} - \frac{p_{n - 1}}{q_{n - 1}} = 0$$ so that the infinite continued fraction is convergent.

Hence we have the following result:
If the series $\sum a_{n - 1}a_{n} / b_{n}$ is divergent then the corresponding infinite continued fraction of the first class is convergent and its value is a positive real number.

To handle the infinite continued fractions of the second class we make the following observations (the proofs are exactly same as that provided for the continued fractions of the first class):
For an infinite continued fraction of the second class, subject to the condition $a_{n} \geq b_{n} + 1$, all the convergents are positive and form a strictly increasing sequence.

Hence in this case there is a chance that the continued fraction can diverge to $\infty$. We also need to understand that the divergence here is actually somewhat accidental in the sense that if we remove some terms of the continued fraction from the beginning it may be convergent. To explain further why this is a possibility assume that the continued fraction formed after removing first 4 terms (i.e. starting the fraction from $a_{5}, b_{5}$) is convergent. It might just happen that the value of this infinite continued fraction comes out to be $a_{4}$ and then the continued fraction starting with $a_{4}, b_{4}$ is divergent and that starting with $a_{3}, b_{3}$ converges to $0$. Hence in this case the convergence or divergence of a continued fraction is heavily dependent on each term forming the infinite continued fraction. This is in quite contrast with the case of an infinite series of a product where a finite number of terms can always be removed without affecting the property of convergence or divergence. Also note that this is quite a different behavior from the continued fractions of the first class.

To obtain a criterion of convergence for the continued fractions of the second class, it needs to be provided in a different form and then we have the following result:
If an infinite continued fraction of the second class of the form
$$\frac{b_{2}}{a_{2} - }\,\frac{b_{3}}{a_{3} - }\,\frac{b_{4}}{a_{4} - }\,\frac{b_{5}}{a_{5} - }\,\cdots$$ be such that $a_{n} \geq b_{n} + 1$ for all values of $n$, then the infinite continued fraction converges to value $A$ with $0 < A \leq 1$. If $a_{n} > b_{n} + 1$ at least once then $0 < A < 1$. If $a_{n} = b_{n} + 1$ for all values of $n$ then $A = 1 - (1/B)$ where $$B = 1 + b_{2} + b_{2}b_{3} + b_{2}b_{3}b_{4} + \cdots + b_{2}b_{3}\cdots b_{n} + \cdots$$ and therefore the value $A < 1$ or $A = 1$ according as whether the above infinite series is convergent or divergent.

To prove this we need to observe that \begin{align} p_{n} - p_{n - 1} &= (a_{n} - 1)p_{n - 1} - b_{n}p_{n - 2} \geq b_{n}(p_{n - 1} - p_{n - 2})\notag\\ p_{n - 1} - p_{n - 2} &\geq b_{n - 1}(p_{n - 2} - p_{n - 3})\notag\\ \cdots &\geq \cdots\notag\\ p_{3} - p_{2} &\geq b_{3}b_{2}\notag \end{align} Multiplying these and noting that both sides of each inequality are positive we get $$p_{n} - p_{n - 1} \geq b_{2}b_{3}\cdots b_{n}$$ and using this equation for $n = 2, 3, \ldots, n$ and on adding these we get $$p_{n} \geq b_{2} + b_{2}b_{3} + \cdots + b_{2}b_{3}\cdots b_{n}$$ Similarly we can establish $$q_{n} \geq 1 + b_{2} + b_{2}b_{3} + \cdots + b_{2}b_{3}\cdots b_{n}$$ Now it is easy to see that \begin{align} q_{n} - p_{n} &= a_{n}(q_{n - 1} - p_{n - 1}) - b_{n}(q_{n - 2} - p_{n - 2})\notag\\ &\geq (b_{n} + 1)(q_{n - 1} - p_{n - 1}) - b_{n}(q_{n - 2} - p_{n - 2})\text{ provided }q_{n - 1} - p_{n - 1} > 0\notag\\ &= (q_{n - 1} - p_{n - 1}) + b_{n}\{(q_{n - 1} - p_{n - 1}) - (q_{n - 2} - p_{n - 2})\}\notag \end{align} Now $q_{2} - p_{2} = a_{2} - b_{2} \geq 1$ and \begin{align} q_{3} - p_{3} &= a_{2}a_{3} - b_{3} - b_{2}a_{3}\notag\\ &\geq (a_{2} - b_{2})(b_{3} + 1) - b_{3}\notag\\ &\geq (a_{2} - b_{2}) + b_{3}(a_{2} - b_{2} - 1)\notag\\ &\geq a_{2} - b_{2} = q_{2} - p_{2}\notag \end{align} It now follows by successive application of the same argument that $$q_{n} - p_{n} \geq q_{n - 1} - p_{n - 1} \geq q_{n - 2} - p_{n - 2} \geq \cdots \geq q_{2} - p_{2} \geq 1$$ And therefore we have $p_{n}/q_{n} \leq 1 - (1/q_{n})$ and since $q_{n} > 1$, it follows that the sequence of convergents is bounded above and since it is an increasing sequence, it follows that the sequence of convergents has a limit $A$ with $0 < A \leq 1$. Now if the condition $a_{n} > b_{n} + 1$ holds for at least one value $n = m$ i.e. $a_{m} > b_{m} + 1$, then we have $0 < A < 1$. To see that this is the case let us define $$A_{i} = \frac{b_{i}}{a_{i} -}\,\frac{b_{i + 1}}{a_{i + 1} -}\,\frac{b_{i + 2}}{a_{i + 2} -}\,\frac{b_{i + 3}}{a_{i + 3} -}\,\frac{b_{i + 4}}{a_{i + 4} -}\,\cdots$$ From the argument above we have $0 < A_{i} \leq 1$ and in particular $0 < A_{m + 1} \leq 1$. Then we have $$A_{m} = \frac{b_{m}}{a_{m} - A_{m + 1}} \leq \frac{b_{m}}{a_{m} - 1} < 1$$ and so we have $$A_{m - 1} = \frac{b_{m - 1}}{a_{m - 1} - A_{m}} < \frac{b_{m - 1}}{a_{m - 1} - 1} \leq 1$$ so that $A_{m - 1} < 1$. Continuing backwards in this fashion we can see that $0 < A < 1$.

Moreover if $a_{n} = b_{n} + 1$ for all values of $n$ then $$q_{n} - p_{n} = q_{n - 1} - p_{n - 1} = q_{n - 2} - p_{n - 2} = \cdots = q_{2} - p_{2} = 1$$ so that $p_{n} / q_{n} = 1 - (1/q_{n})$ and since $$q_{n} = 1 + b_{2} + b_{2}b_{3} + \cdots + b_{2}b_{3}\cdots b_{n}$$ it follows that the limit $A = \lim_{n \to \infty} p_{n} / q_{n}$ is less than unity or equal to unity according as the series above is convergent or divergent.

The reader can apply these tests of convergence on the infinite continued fractions of $\tan x$ and $\tanh x$ obtained in an earlier post and thereby provide a full justification of these infinite continued fractions.

We will deal with the issues of irrationality of continued fractions in the next post and thereby present the Lambert's original proof of irrationality of $\pi$.

Note: Content of this and the following post are derived from Chrystal's Algebra Vol. 2 and the reader is encouraged to read this famous and slightly old-style book.