How to prove that $\frac{\pi}{2}=\left[1, 1, \tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\right]$?

The article shows that as the harmonic series $\sum\frac{1}{n}$ diverges, one can apply Van Vleck's theorem to show that the $2n$th and $(2n+1)$th convergents of the continued fraction $[1; 1, 1/2, 1/3,\dots]$ both exist and converge to the same limit.

The author transforms $[1;1,1/2,1/3,\dots]$ into the form $$1+\frac{1\cdot1}{1+\frac{1\cdot2}{1+\frac{2\cdot3}{1+\dots}}}$$ by multiplying the numberator and the denominator of the $n$th fraction by $n$. Let $f_n$ be the numerator or the denominator of the $n$th (even) convergent then $f_n$ satisfies the relation $$f_n=(2n^2-4n+3)f_{n-2}-(n-1)(n-2)^2(n-3)f_{n-4}$$ By direct substitutions, one can show that $A_n$ and $B_n$ where $A_n=\prod_{k\text{ even}}^nk^2$ and $B_n=(n+1)\prod_{k\text{ odd}}^{n-1}k^2$ satisfy the relation for even $n$s. The fraction $\frac{A_n}{B_n}$ has a limit known as the Wallis product. Since for odd $n$s the $n$th convergent converges to the same limit as previously proved, the continued fraction converges to $\frac{\pi}{2}$.


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Edit:

(1) If you are not familiar with the Van Vleck theorem, it is helpful to study the Seidel-Stern theorem. This theorem states that if $\sum b_n=\infty$ and $b_n$ positive, then $K(1/b_n)$ (the continued fraction) converges. The proof of this theorem can be found in Waadeland and Lorentzen's text Continued Fractions with Applications, Chapter 3 Theorem 3. The proof takes advantage of the fact that $$\frac{A_{2n+1}}{B_{2n+1}}-\frac{A_{2n}}{B_{2n}}=\frac{1}{B_{2n}B_{2n+1}}\to 0$$ as $n\to\infty$. This can be seen from the recursive relation $$B_{n}=b_nB_{n-1}+B_{n-2}$$ Since the $b_n$ are positive, we have $B_{2n}>B_{2n-2}>\cdots>B_0=1$ and $B_{2n+1}>\cdots>B_1=b_1$. Therefore $$B_{2n}>b_{2n}b_1+B_{2n-1}>\cdots>(b_{2n}+b_{2n-2}+\cdots+b_2)b_1+1$$ and $$B_{2n+1}>b_{2n+1}+B_{2n-1}>\cdots>b_{2n+1}+b_{2n-1}+\cdots+b_1$$ Since $\sum b_n$ divergese, the denominator goes to infinity with $n$.

Van Vleck's theorem, on the other hand, concerns the convergence of the continued fraction $K(a_n/b_n)$ with complex $a_n, b_n$. It gives a more general criterion for convergence according to the argument of $b_n$ and the divergence of $\sum |b_n|$ (in our case the argument is zero and $|b_n|=b_n$).

(2) The recursive relation is in no way magical. Let $f_n$ be the denominator or the numerator of the $n$th convergent, we have $$f_n=b_nf_{n-1}+a_nf_{n-2}$$ $$f_{n-1}=b_{n-1}f_{n-2}+a_{n-1}f_{n-3}$$ $$f_{n-2}=b_{n-2}f_{n-3}+a_{n-2}f_{n-4}$$ like any continued fraction. Then $f_{n-3}=1/b_{n-2}(f_{n-2}-a_{n-2}f_{n-4})$. Plugging the second relation into the first one, one gets $$f_n=(b_{n}b_{n-1}+a_n+a_{n-1}b_n/b_{n-2})f_{n-2}-a_{n-1}a_{n-2}b_n/b_{n-2}f_{n-4}$$

After the transformation we did, $b_i=1$ for all $i$ and $a_n=n(n-1)$ for $n>1$. Thus, $$\begin{aligned} f_n &=(1+n(n-1)+(n-2)(n-1))f_{n-2}-(n-1)(n-2)^2(n-3)f_{n-4}\\ &=(2n^2-4n+3)f_{n-2}-(n-1)(n-2)^2(n-3)f_{n-4} \end{aligned}$$


References:

Lorentzen, L. and Waadeland, H. (1992). Continued fractions with applications. Amsterdam ; London ; New York ; Tokyo: Elsevier Science Publishers B. V.

Pickett, T.J. and Coleman, A. (2008). Another Continued Fraction for π. The American Mathematical Monthly, 115(10), pp.930–933.

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Matrix Representation of Convergents

One method to compute the convervgents of continued fractions uses matrices: if $[c_0;c_1,c_2,\dots,c_n]=\frac{p_n}{q_n}$, then $$ \begin{bmatrix}0&1\\1&0\end{bmatrix} \prod_{k=0}^n\begin{bmatrix}0&1\\1&c_k\end{bmatrix} =\begin{bmatrix}p_{n-1}&p_n\\q_{n-1}&q_n\end{bmatrix}\tag1 $$ Since $c_0=1$, and for $k\ge1$, $c_k=\frac1k$ we get $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} \prod_{k=1}^n\begin{bmatrix}0&1\\1&\frac1k\end{bmatrix} =\begin{bmatrix}p_{n-1}&p_n\\q_{n-1}&q_n\end{bmatrix}\tag2 $$


Closed Form

We will show by induction that $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} \prod_{k=1}^{2n}\begin{bmatrix}0&1\\1&\frac1k\end{bmatrix} =\begin{bmatrix} \frac{4^n}{\binom{2n}{n}}&\frac{4^n}{\binom{2n}{n}}\\ \frac{\binom{2n}{n}2n}{4^n}&\frac{\binom{2n}{n}(2n+1)}{4^n} \end{bmatrix}\tag3 $$ Note that $(3)$ is true for $n=0$. Assume $(3)$ is true for some $n$, then $$ \begin{bmatrix} \color{#C00}{\frac{4^n}{\binom{2n}{n}}}&\color{#C00}{\frac{4^n}{\binom{2n}{n}}}\\ \color{#090}{\frac{\binom{2n}{n}2n}{4^n}}&\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}} \end{bmatrix} \begin{bmatrix} 0&1\\1&\frac1{2n+1} \end{bmatrix} =\begin{bmatrix} \frac{4^n}{\binom{2n}{n}}&\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\\ \frac{\binom{2n}{n}(2n+1)}{4^n}&\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}} \end{bmatrix}\tag4 $$ The left column of the result is easy, the right column top element is $$ \begin{align} \color{#C00}{\frac{4^n}{\binom{2n}{n}}}\cdot1+\color{#C00}{\frac{4^n}{\binom{2n}{n}}}\cdot\frac1{2n+1} &=\frac{4^n}{\binom{2n}{n}}\frac{2n+2}{2n+1}\tag{5a}\\ &=\frac{4^n}{\binom{2n}{n}}\frac{(2n+2)^2}{(2n+1)(2n+2)}\tag{5b}\\ &=\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\tag{5c} \end{align} $$ the right column bottom element is $$ \begin{align} \color{#090}{\frac{\binom{2n}{n}2n}{4^n}}\cdot1+\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\cdot\frac1{2n+1} &=\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\tag6 \end{align} $$ We can continue $$ \begin{bmatrix} \color{#C00}{\frac{4^n}{\binom{2n}{n}}}&\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\\ \color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}&\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}} \end{bmatrix} \begin{bmatrix} 0&1\\1&\frac1{2n+2} \end{bmatrix} =\begin{bmatrix} \frac{4^{n+1}}{\binom{2n+2}{n+1}}&\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\\ \frac{\binom{2n}{n}(2n+1)}{4^n}&\color{#090}{\frac{\binom{2n+2}{n+1}(2n+3)}{4^{n+1}}} \end{bmatrix}\tag7 $$ The left column of the result is easy, the right column top element is $$ \begin{align} \color{#C00}{\frac{4^n}{\binom{2n}{n}}}\cdot1+\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\cdot\frac1{2n+2} &=\frac{4^n}{\binom{2n}{n}}+\frac{4^n}{\binom{2n}{n}}\frac4{\frac{(2n+1)(2n+2)^2}{(n+1)^2}}\tag{8a}\\ &=\frac{4^n}{\binom{2n}{n}}+\frac{4^n}{\binom{2n}{n}}\frac1{2n+1}\tag{8b}\\ &=\frac{4^n}{\binom{2n}{n}}\frac1{\frac{(2n+1)(2n+2)}{(2n+2)^2}}\tag{8c}\\ &=\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\tag{8d} \end{align} $$ the right column bottom element is $$ \begin{align} \color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\cdot1+\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\cdot\frac1{2n+2} &=\frac{\binom{2n}{n}(2n+1)}{4^n}\frac{(2n+2)(2n+3)}{(2n+2)^2}\tag{9a}\\ &=\color{#090}{\frac{\binom{2n+2}{n+1}(2n+3)}{4^{n+1}}}\tag{9b} \end{align} $$ $(4)$ and $(7)$ show that $(3)$ is true for $n+1$.


Estimating the Convergents

$(1)$ and $(3)$ show that $$ \frac{p_{2n-1}}{q_{2n-1}}=\frac1{2n}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\tag{10} $$ $$ \frac{p_{2n}}{q_{2n}}=\frac1{2n+1}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\tag{11} $$ $(10)$ and $(11)$ can be unified as $$ \frac{p_n}{q_n}=\frac1{n+1}\left(\frac{4^{\lceil n/2\rceil}}{\binom{2\lceil n/2\rceil}{\lceil n/2\rceil}}\right)^2\tag{12} $$ $(9)$ from this answer says that $$ \pi\left(n+\tfrac14\right)\le\left(\frac{4^n}{\binom{2n}{n}}\right)^2\le\pi\left(n+\tfrac13\right)\tag{13} $$ Since $\frac n2\le\lceil n/2\rceil\le\frac{n+1}2$, $(12)$ and $(13)$ show that $$ \frac\pi2\left(1-\frac1{2n+2}\right)\le\frac{p_n}{q_n}\le\frac\pi2\left(1+\frac2{3n+3}\right)\tag{14} $$ which, by the Squeeze Theorem, says that $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\left[1;1,\tfrac12,\tfrac13,\dots,\tfrac1n\right]=\frac\pi2}\tag{15} $$ enter image description here


Wallis Product

The Wallis Product for $\pi$ is $$ \frac\pi2=\prod_{k=1}^\infty\frac{4k^2}{4k^2-1}\tag{16} $$ Consider the partial products $$ \begin{align} \prod_{k=1}^n\frac{4k^2}{4k^2-1} &=\prod_{k=1}^n\frac{2k}{2k+1}\frac{2k}{2k-1}\tag{17a}\\ &=\prod_{k=1}^n\frac{2k-1}{2k+1}\left(\frac{2k}{2k-1}\right)^2\tag{17b}\\ &=\frac1{2n+1}\left(\prod_{k=1}^n\frac{(2k)^2}{(2k-1)2k}\right)^2\tag{17c}\\ &=\frac1{2n+1}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\tag{17d} \end{align} $$ Note the similarity between $(11)$ and $\text{(17d)}$.