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Can someone give me a hint on how to prove the following claim? I don't know how to start the proof.

$$\dfrac{\pi}{4}=1+ \cfrac{1}{3- \cfrac{3\cdot 4}{1- \cfrac{2 \cdot 3}{3- \cfrac{5 \cdot 6}{1-\cfrac{4 \cdot 5}{\ddots}}}}}$$

PARI/GP code.

Pedja
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  • Some references are given in Wikipedia Pi # continued fraction, but apparently not for this one. There are also some threads on this topic here, at MSE, like How to find continued fraction of pi or A remarkable Continued Fraction for π and many more. – CiaPan Jan 15 '25 at 13:42
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    What's the generic pattern here? That should be in the post itself, not in an external link. – Sil Jan 15 '25 at 14:10
  • If I was a math student assigned this problem, my first question would be: what is the intent of the problem composer? Normally the composer intends the problem to be an application of math tools that you have recently been exposed to (i.e. worked examples, proven theorems, previously solved problems). So, I would try to identify and apply the corresponding tools. – user2661923 Jan 15 '25 at 14:51
  • For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. – user2661923 Jan 15 '25 at 14:51
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    Just a thought: If you take $\tan$ of both sides, and use the sum of angles formula for $\tan$, then by induction, you might be able to cook up a telescoping sum...? – B. Goddard Jan 15 '25 at 17:08
  • The value of a continued fraction is the limit of the sequence formed by removing the second term in the denominators. That is $$a_1 = 1\a_2 = 1+\frac 13\a_3 = 1 + \dfrac 1{3 - \frac{3\cdot 4}1}\\vdots$$ Calculate several terms of this sequence. See if you can deduce a pattern. See if you can at least show it is increasing and bounded above by $1$. Thus it converges. Then see if you can show $\frac \pi 4$ is the least upper bound. – Paul Sinclair Jan 17 '25 at 12:38

1 Answers1

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Stern (1833)$^\dagger$ derived the following relation between infinite products and infinite continued fractions (by transforming the product into a series, then converting the series into a continued fraction): $${d_1\over e_1}\cdot{d_2\over e_2}\cdot{d_3\over e_3}\cdots\ \ =\\[3ex] 1+{d_1-e_1\over e_1-}\ {(d_2-e_2)d_1e_1\over(d_1d_2-e_1e_2)-}\ {(d_1-e_1)(d_3-e_3)d_2e_2\over(d_2d_3-e_2e_3)-}\dots\ {(d_{m-1}-e_{m-1})(d_{m+1}-e_{m+1})d_me_m\over(d_md_{m+1}-e_me_{m+1})-}\dots$$ where on the RHS I've used the common notation for infinite continued fractions $$b_0\pm{a_1\over b_1\pm}\ {a_2\over b_2\pm}\ {a_3\over b_3\pm}\cdots\ \ :=\ \ b_0 \pm \cfrac{a_1}{b_1 \pm \cfrac{a_2}{b_2 \pm \cfrac{a_3}{b_3 \pm \ddots\,}}}$$

As an example application, Stern wrote the Wallis product as $${\pi\over 2}=\frac{2}{3} \cdot \frac{2}{1} \cdot \frac{4}{5} \cdot \frac{4}{3} \cdot \frac{6}{7} \cdot \frac{6}{5} \cdots $$ from which the above relation gives the continued fraction $${\pi\over 2}=1-{1\over3-}\ {2\cdot3\over1+}\ {1\cdot2\over3+}\ {4\cdot5\over1+}\ {3\cdot4\over3+}\ {6\cdot7\over1+}\cdots $$ However, we can also write $${\pi\over 4}=\frac{4}{3} \cdot \frac{2}{3} \cdot \frac{6}{5} \cdot \frac{4}{5} \cdot \frac{8}{7} \cdot \frac{6}{7} \cdots$$ from which the relation yields $${\pi\over 4}=1+{1\over3-}\ {3\cdot4\over1-}\ {2\cdot3\over3-}\ {5\cdot6\over1-}\ {4\cdot5\over3-}\ {7\cdot8\over1-}\cdots $$

$^\dagger$ M. Stern. "Theorie der Kettenbrüche und ihre Anwendung. (Fortsetzung)." Journal für die reine und angewandte Mathematik 10 (1833): 241-274. Free access (in German). See esp. pp.266ff. Note that Stern's notation for our $b_0\pm{a_1\over b_1\pm}\ {a_2\over b_2\pm}\ {a_3\over b_3\pm}\cdots\ \ $ is $\ \ F(b_0\pm a_1:b_1\pm a_2:b_2\pm\cdots).$

Another source for this is Chrystal's Algebra: an elementary text-book, Part II (1906), which carries out a similar development on pp.515-517, and cites Stern (1833); however, Chrystal chooses to write the result in the (equivalent) form
$${d_1\over e_1}\cdot{d_2\over e_2}\cdot{d_3\over e_3}\cdots\ \ =\\[3ex]{d_1\over e_1-}\ {(d_2-e_2)e_1\over d_2-}\ {(d_3-e_3)d_2e_2\over(d_2d_3-e_2e_3)-}\ {(d_2-e_2)(d_4-e_4)d_3e_3\over(d_3d_4-e_3e_4)-}\dots\ {(d_{m-1}-e_{m-1})(d_{m+1}-e_{m+1})d_me_m\over(d_md_{m+1}-e_me_{m+1})-}\dots $$ which is less suited to obtaining the OP's continued fraction.

r.e.s.
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