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

I don't know how to prove the generalized continued fraction $$\pi/2=[1;1/1,1/2,1/3,1/4,...]=1+\cfrac{1}{1/1+\cfrac{1}{1/2+\cfrac{1}{1/3+\cfrac{1}{1/4+\ddots}}}}$$

It appears on wikipedia without proof, it contains the terms of the harmonic series but it seems not to be very well known.

I would appreciate any help, thanks.

• The reference seems to be from maa.org/press/periodicals/american-mathematical-monthly/… ... Pickett and Coleman, "Another Continued Fraction for $\pi$." But the full text appears to require a subscription. Commented Dec 20, 2020 at 22:41
• From that article it is noted that the convergents of this continued fraction $a_n = [1; 1,1/2,\ldots,1/n]$ gives the partial product of the Wallis product en.wikipedia.org/wiki/Wallis_product (1, 2/1, 4/3, 16/9,...). Perhaps you can try to prove that (issue of convergence aside...). Commented Dec 20, 2020 at 22:54
• @mjqxxxx thanks, I don't have access either so I will try to see what I can do but it really help to have a reference. Commented Dec 21, 2020 at 19:13
• @bonsoon thanks, yes that is the part I need to understand the most, Yourong 'DZR' Zang answer mentions Van Vleck's theorem to this part so I'm trying to read the wikipedia page to get the idea of how it goes. Commented Dec 21, 2020 at 19:13
• Added the given reference and another one to the wikipedia page Commented Jan 8, 2021 at 22:26

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 $$\frac{\pi}{2}$$.

I hope this is a clear summary of the article that can save you some time. Sorry for falsely flagging your question.

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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• Sorry for the delay, I have upvoted but will wait for others answers for now before accepting, I didn't notice the flag but worry not, moderation is as needed as answers so I appreciate both, I have read about the Wallis product before but I'm not really used to continued fractions and I had never heard of the Van Vleck's theorem, the recurrence relation looks very magical so I'm trying to read the wikipedia page of the theorem to get more sense of how it appears. Commented Dec 21, 2020 at 19:14
• Hi @DanielD. I have updated my answer. I have provided a more elementary convergence criterion that does not require any background knowledge of continued fractions (except for the simplest recursive relation). I have copied the proof of the recursive relation which you found magical from the article. I hope the new version would provide you with a more acceptable explanation.
– user672528
Commented Dec 22, 2020 at 9:15
• @DanielD. Also, $f_n$ in the recursion formula is the not the $n$th convergent. Instead, it is the denominator ($B_n$) or the numerator ($A_n$) of the $n$th convergent. In the article, the author denotes by $f$ the $n$th convergent, and uses $f$ for the recursion, so it's pretty confusing.
– user672528
Commented Dec 22, 2020 at 9:37
• Another nice proof of Seidel-Stern can be found in Khinchin's book: Continued Fractions. I say "another" proof, not to suggest that it's different in some essential way from what's sketched here, but simply to provide a reference to another author writing it up. Commented Dec 22, 2020 at 10:02
• @DanielD. I'm writing up the induction argument as I remember working it out a few years ago. It's probably inferior to this answer here is many ways, but it might be more accessible for some, and I enjoy the exercise of remembering the details. Not sure when I'll actually hit 'post', though, as I'm simultaneously setting up birthday surprises for a brand new 3-year-old :) Commented Dec 22, 2020 at 12:52

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 $$\frac{p_n}{q_n}=\frac1{n+1}\left(\frac{4^{\lceil n/2\rceil}}{\binom{2\lceil n/2\rceil}{\lceil n/2\rceil}}\right)^2\tag{12}$$ $$(10)$$ 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 $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\left[1;1,\tfrac12,\tfrac13,\dots,\tfrac1n\right]=\frac\pi2}\tag{15}$$

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 $$\text{(17d)}$$.

• Pretty neat proof! I like how the squeeze theorem shows how useful is to separate odd and even terms, I put some terms on wolfram and try also changing 1/2,1/3 by 1/k+1,1/k+3 for example and other things but I can't see how you realized the fornula for the 2n matrix Commented Dec 23, 2020 at 16:24
• and in the link you use that $(\frac{n+\frac12}{n+1})^2 \le\frac{n+\frac13}{n+\frac43}$ which cross multiplying and resting terms checks out but even though it looks simple I don't have idea either how you realized this was the inequality you needed Commented Dec 23, 2020 at 16:32
• The entries in the $2\times2$ matrix were deduced from looking at the values of $p_n$ and $q_n$ in the convergents of the continued fraction and looking at the values of Wallis' Product.
– robjohn
Commented Dec 23, 2020 at 16:43
• In the link, what one wants is to get a telescoping product, that is where the numerator and denominator differ by $1$, as in $\frac{n+1/3}{n+4/3}$. Since the numerator and denominator in $\frac{n+1/2}{n+1}$ differ by $\frac12$, squaring is indicated. I just needed to figure out that $$\frac{n+1/4}{n+5/4}\le\left(\frac{n+1/2}{n+1}\right)^2\le\frac{n+1/3}{n+4/3}$$
– robjohn
Commented Dec 23, 2020 at 16:52
• Thanks for your edit it really helped me understand what I couldn't and I wouldn't have realized either the Wallis product was behind this answer as well if you didn't mentioned it, great visualization too, I surely will come back to let the answers sink more on my head Commented Dec 25, 2020 at 19:27