Lambert's Original Proof that $\pi$ is irrational. I am trying to find Lambert's original proof that $\pi$ is irrational. Wikipedia has a little description but it is quite lacking. Can someone direct me to Lambert's original proof or post his proof here?
Thanks.
 A: Here is the complete proof:
Part I Derivation of continued fraction for $\tan(x)$
Part II Proof of Irrationality of continued fractions
Part III Proof of Irrationality of $\pi$.
Part I
We show 
$$\tan x=\cfrac{x}{1-
\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$
First we treat Lambert's continued fraction expression for the quotient of two power series.
Let $F_0$ and $F_1$ be two power series we may assume without loss of generality that they both have nonzero constant term
and that in fact  both have constant term of $1$.
Define
$$F_1-F_0=b_1 x F_2$$
$$F_2-F_1=b_2 x F_3$$
$$F_3-F_2=b_3 x F_4$$
$$F_{n+1}-F_n =b_{n+1} x F_{n+2}$$
where at each step we make the assumption that $F_n$ has a constant term of $1$. In general we would simply get higher powers of $x$ in the equation.
Define $$G_n=\frac{F_{n+1}}{F_n}$$ then we have by dividing the last equation by $F_{n+1}$,
$$G_n=\frac{1}{1-b_{n+1}xG_{n+1}}$$
therefore
$$\frac{F_1}{F_0}=G_0=\cfrac{1}{1-
\cfrac{b_1 x}{1-\cfrac{b_ 2 x}{1-\cfrac{b_3 x}{1-\cdots}}}}$$
Now we shall apply this to obtain continued fraction expansions for several functions including $\tan x$.
Let us define the particular power series,
\begin{equation*}
\begin{split}
F_n =&1+ \frac{x}{1! (\gamma +n)} +\frac{x^2}{2! (\gamma +n)(\gamma +n+1)} +\frac{x^3}{3! (\gamma +n)(\gamma +n+1)(\gamma +n+2)} \cdots \\
&1+ \sum\limits_{k=1}^{\infty} \frac{x^k}{k! (\gamma +n)\cdots (\gamma +n+k-1)}
\end{split}
\end{equation*}
Then we have
$$F_{n+1}-F_n =-\frac{x}{(\gamma +n)(\gamma +n+1)} F_{n+2}.$$
Thus with $b_n =  -\frac{x}{(\gamma +n)(\gamma +n+1)}$ we get
\begin{equation*}
\begin{split}
\frac{F_1}{F_0}=&\cfrac{1}{1+\cfrac{\frac{x}{(\gamma)(\gamma +1)}}{1+\cfrac{\frac{x}{(\gamma +1)(\gamma +2)}}{1+\cfrac{\frac{x}{(\gamma +2)(\gamma +3)}}{1+\cdots}}}} \\
&=\cfrac{\gamma}{\gamma +\cfrac{x}{(\gamma+1)+\cfrac{x}{(\gamma +2)+\cfrac{x}{(\gamma +3)+\cdots}}}}
\end{split}
\end{equation*}
Now let us set $\gamma=\frac{1}{2}$ and instead of $x$ write $-\frac{x^2}{4}$
Then $$F_1= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k+1)!}=\frac{\sin x}{x}$$
and
$$F_0= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}=\cos x.$$
Putting this altogether we get
$$\tan x=\cfrac{x}{1-
\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$
Part II
In the continued fraction,
$$\cfrac{b_1}{a_1 -
\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}}}  $$
assume that
where $1+b_n \leq a_n$ for all $n$, and that we have $1+b_n < a_n$ infinitely often.
Then the fraction is irrational.
Assume that the fraction is rational, say
$$\frac{\lambda_1}{\lambda_0}=\cfrac{b_1}{a_1 -
\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}}} $$
where $\lambda_1$ and $\lambda_0$ are positive integers, now since the fraction converges to a number less than one, $\lambda_1 < \lambda_0$
If we set $$\rho_1=\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}} $$
then we have
$$\frac{\lambda_1}{\lambda_0}=\frac{b_1}{a_1 -\rho_1}$$
so$$\rho_1=\frac{a_1 \lambda_1 - b_1 \lambda_0}{\lambda_1} < 1$$
So $\rho_1=\frac{\lambda_2}{\lambda_1}$ where $\lambda_2 < \lambda_1$.
Continuing in this way we obtain a strictly decreasing sequence of positive integers, $\lambda_0 > \lambda_1 > \cdots$, a contradiction.
Part III
This expression leads to the following fundamental result.
$\pi$ is irrational.
Assume that $\pi$ is rational, then $\frac{\pi}{4}$ is also rational.
Let $$\frac{\pi}{4}=\frac{a}{b},$$
and substitute $x=\frac{\pi}{4}$ into Lambert's continued fraction for $\tan x$.
We get
\begin{equation*}
\begin{split}
1=&\cfrac{\frac{a}{b}}{1-
\cfrac{\frac{a^2}{b^2 }}{3-\cfrac{\frac{a^2 }{b^2}}{5-\cfrac{\frac{a^2}{b^2}}{7-\cdots}}}} \\
&=\cfrac{a}{b-
\cfrac{a^2}{3b-\cfrac{a^2}{5b-\cfrac{a^2}{7b-\cdots}}}} \\
\end{split}
\end{equation*}
 Now since eventually $nb > a^2 +1$ we have that this expression is irrational, and this is absurd since it is equal to $1$.
 Therefore $\pi$ is irrational.
A: There is a proof here, but i am not sure if it's helpful or not: http://www.pi314.net/eng/lambert.php
Here is a link to the original article: http://www.kuttaka.org/~JHL/L1768b.pdf
A: This post is meant to supplement the useful answer of Rene Schipperus, in view of the unanswered comments by Isomorphism and templatetypedef under Schipperus's answer (which seem to me to be valid concerns regarding Part II).
Claim: let $a$ be a positive integer, and let $b_1, b_2, b_3, \ldots$ be a sequence of positive integers, and assume that for all $n$ we have
$$0 < \cfrac{a}{b_n - \cfrac{a}{b_{n+1} - \cfrac{a}{b_{n+2} - \ldots}}} < 1.$$
Then each of these continued fractions is irrational.
Putting aside questions of convergence, the proof was essentially given by Schipperus in his answer. Without loss of generality, we prove the continued fraction in the case $n=1$ is irrational. Suppose it were rational, say $\frac{c_2}{c_1}$ where $c_1, c_2$ are positive integers. Then $c_1 > c_2$ by our assumption in the case $n=1$. Write
$$\frac{c_2}{c_1} = \frac{a}{b_1 - e_2}$$
so that $e_2$ is the continued fraction above in the case $n=2$. By our assumption, we have $0 < e_2 < 1$. Solving for $e_2$, we have
$$e_2 = \frac{c_2b_1 - ac_1}{c_2}$$
so that defining $c_3 = c_2b_1 - ac_1$, we have that $c_3$ is a positive integer less than $c_2$. Proceeding in this way, we produce a strictly decreasing sequence of positive integers $c_1 > c_2 > c_3 > \ldots$, contradiction.
Now we finish off Schipperus's proof as follows. Consider the sequence of numbers
$$\cfrac{a^2}{(2k+1)b - \cfrac{a^2}{(2k+3)b - \cfrac{a^2}{(2k+5)b - \ldots}}}$$
for some large $k$. It is enough to prove that one of these numbers is irrational. But clearly these numbers decrease as $k$ increases, and for large enough $n$ we reach a point where for all $k \geq n$, all these numbers are in the range strictly between $0$ and $1$. This completes the proof.
