Infinite products of Bernoulli and Euler

In 1720, Christian Goldbach's experiments with infinite series was published in the German journal Acta Eruditorum, he titled his paper Specimen methodi ad summas serierum and proceeded to furnish a number of worked examplesThe following examples are given in Goldbach's 1720 paper: $$\begin{align}\tfrac{1}{2}+\tfrac{1}{6}+\tfrac{1}{12}+\tfrac{1}{20}+\ldots\\ \tfrac{2}{64}+\tfrac{11}{544}+\tfrac{26}{1972}+\tfrac{47}{5104}+\ldots \\ \tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots \\ \tfrac{1}{3}+\tfrac{1}{81}+\tfrac{1}{19683}+\tfrac{1}{43046721} + \ldots\end{align}$$ whose general terms are $\frac{1}{x^2+x}$, $\frac{3x^2-1}{9x^4+36x^3+33x^2-6x-8}$, $\frac{1}{2x}$, and $\frac{1}{2x^x-x}$. of his finite difference technique for relocating the general term of a given sequence. The general principle is that in the general case, the $n$-th term, $a_n$, of any number sequence $$a_1, a_2, a_3, \ldots$$ can be expressed by using an infinite series involving differences of known terms: $$a_n = a_1 + \frac{n-1}{1!}\Delta a_1 + \frac{(n-1)(n-2)}{2!}\Delta^2 a_1 + \ldots$$ where the delta's are iteratively defined by $\Delta a_1 = a_{n+1} - a_n$ and $$\Delta^n a_1 = \Delta^{n-1}a_{n+1} - \Delta^{n-1}a_n=\sum_{k=0}^{n-1}\binom{n-1}{k}a_{n+1-k}$$ This recurrence is actually related to the backward finite difference scheme, since the modern rewrite of the equation involves binomial coefficients.

Goldbach recounted, in his letter to Daniel Bernoulli, that he was introduced to the subject of infinite series by Daniel's uncle, Nikolaus I at Oxford in 1712. Nikolaus gave Goldbach some reading materials (written by Nikolaus's older brother Jakob) but Goldbach was not able to fully comprehend the work. His interest in the subject was refreshed a few years later when he read Leibniz's article in Acta eruditorum in 1717, which motivated him to submit his own infinite series entry to the same journal in 1720. In modern notation, $\Delta^n a_1$ can be written as: $\Delta^n a_1=\sum_{k=0}^{n-1}\binom{n-1}{k}a_{n+1-k}$.

Construction of the general term is curiously important because you can then use the formula to inject an arbitary members between terms in the original number sequence. For example, we can ask: what should be the value assigned to $\color{brown}{a_{\frac{1}{2}}}$, $\color{brown}{a_{1\frac{1}{2}}}$, $\color{brown}{a_{2\frac{1}{2}}}$, etc? $$\phantom{.}^{\searrow} \color{brown}{a_{\frac{1}{2}}} , \quad a_1, \phantom{.}^{\searrow} \color{brown}{a_{1\frac{1}{2}}}, \quad a_2, \phantom{.}^{\searrow}\color{brown}{a_{2\frac{1}{2}}}, \quad a_3, \ldots$$ Many years later in 1728, Goldbach polished the same idea and wrote a paper titled De terminus generalibus serierum. For example, when $a_n = \frac{1}{n!}$, the equation reduced to $$\tfrac{1}{n!} = 1 - \tfrac{1}{2!}\tfrac{n-1}{1!} + \tfrac{1}{3!}\tfrac{(n-1)(n-2)}{2!} + \tfrac{1}{4!}\tfrac{(n-1)(n-2)(n-3)}{3!} + \ldots$$ and when $n = 1\frac{1}{2}$, $${1\tfrac{1}{2}!} = \frac{1}{1 - \tfrac{1}{2!}\tfrac{1}{2} - \tfrac{1}{3!}\tfrac{1}{2 \cdot 4} + \tfrac{1}{4!} \tfrac{1}{2\cdot 4 \cdot 6} + \ldots}$$ Daniel BernoulliGoldbach met Daniel's brother, Nikolaus, in Venice, and at Nikolaus's suggestion, Goldbach, in April 1723, initiated a series of mail exchanges with Daniel, which lasted until 1730. Daniel's father, Johann, was Euler's teacher at the University of Basel, Switzerland. apparently has a copy of this paper and he presented the Goldbach's idea in a seminar at St. Petersburg Academy, on 23 September 1729. We suppose the young Euler had attended the seminar and he realized that Goldbach's construct of computing the general term of a number sequence using another infinite sequence is not the most elegant solution. Euler must have mentioned this to Daniel, and so Daniel told him to talk to Goldbach directly.

Catherine I (1684-1727) reigned briefly after her husband Peter I died in early 1725. When his grandmother died in mid 1727, the powerful Dolgoruky family moved the 12-year-old Peter II (1715-1730) and the capital from St. Petersburg (St. Peterburg, founded in 1703, is eponymized after Peter I) to Moscow and was later crowned at the Dormition Cathedral on 24 February 1728. About the same time, Goldbach was offered a new job as the tutor to Peter II and his sister, and followed his king to Moscow in January 1728. But Euler, the young men who Goldbach met half-a-year earlier (at the post-burial gatheringSee Footnote 30 in leonhardi euleri opera omnia: In his travel diary Euler notes: “On the 27th the funeral of the late Empress took place . . .we went into the [Peter and Paul] Fortress and the Church, looked on her again and kissed her hand; from there we went to see Mr. de l’Isle and Mr. Leutmann and towards evening visited the Court Physician [Blumentrost], where we met Justice Councillor Goldbach, Mr. Henninger and Mr. Schumacher”. “den 27. war die Leichbegengnuß der seligen Kaiserin . . . da giengen wir in die Festung und Kirche, beschaueten die Kayserin nochmahls und küsseten ihr die Hand, von dar giengen wir zu Mr. de l’Isle und H. Leutman und auf den abend zu H. Leibmedicus, da wir H. Justitien-Raht Goldbach, H. Henninger und H. Schumacher antraffen": Notebook II, quoted according to Mikhailov 1959, p. 278). for Catherine I, on 27 May 1727), stayed remain at St. Petersburg. Goldbach eventually left Moscow and returned to St. Petersburg, when his tutee, Peter II, died unexpectedly in January 1730, when his fiancee Catherine Dolgorukaia infected him with smallpox. See M. Zack, D. Schlimm eds. (2017) Research in History and Philosophy of Mathematics: Proceedings of the 2016 Canadian Society for History and Philosophy of Mathematics in Calgary, Alberta, Birkhauser, Cham, pp. 98-99.

Subsequently in 13This is a date in the old Julian calendar, it is proleptically equivalent to 24 October 1729 (Monday). Later when Euler moved to Berlin, he began to mark his letter with Gregorian dates. However, the dates on Goldbach replies are consistently marked using the Gregorian system. October 1729, the 22-year-old Euler wrote his first mail to Goldbach. This historic mail travelled 700 kilometers southeast from St. Peterburg to Moscow and it carried the following equation: $$m! = \frac{1\cdot 2^m}{\color{brown}{1+m}}\frac{2^{1-m}\cdot3^m}{\color{brown}{2+m}}\frac{3^{1-m}\cdot4^m}{\color{brown}{3+m}}\frac{4^{1-m}\cdot5^m}{\color{brown}{4+m}}\cdots$$ This equation announced to Goldbach that Euler had found a solution to the interpolation problem, mentioned by Daniel Bernoulli in a seminar three weeks earlier.

Euler told Goldbach in the letterFor a modern English translation of the first letter, see M. Mattmüller and Martin, F. Lemmermeyer, eds. (2016) Correspondence of Leonhard Euler with Christian Goldbach Vol. 4A, Birkhäuser, p. 583. The book is also hosted at the University of Basel site., that he knew that Daniel's infinite product was communicated to Goldbach about a week ago. Euler said he showed his infinite product to Daniel and the latter told him that he had a similar-looking result which was used to estimate $1\frac{1}{2}!$. Actually Daniel presented two numerical examples to Goldbach, but his estimate for $1\frac{1}{2}!$ contains a mistake. Daniel wrote in his mail that by taking $x = \frac{3}{2}, A = 8$, he was able to get $1\frac{1}{2}! \approx b(8, 1\frac{1}{2}, 2) \approx 1.3005$. This is apparently incorrect since $\sqrt{\frac{35}{4}}\frac{4}{5}\frac{6}{7}\frac{8}{9}\frac{10}{11}\frac{12}{13}\frac{14}{15}\frac{16}{17}\approx 1.32907$. Daniel must have mis-substituted $3$ for $\frac{3}{2}$ in the surd expression since he wrote $\sqrt{\frac{19}{2}}$ instead of $\sqrt{\frac{35}{4}}$. You can see in this photograph that the value of $1\frac{1}{2}!$ given by Euler is $\frac{3}{4}\sqrt{\pi} = 1.3293403$, a value that was derived not by the infinite product but by a completely new thought process.

After giving some mundane numerical examples on how to use his infinite product formula, e.g. $\frac{4}{3}\frac{9}{8}\frac{16}{15}\frac{25}{24}\frac{36}{35}\cdots = 2$ and $\frac{8}{4}\frac{27}{20}\frac{64}{54}\frac{125}{112} \cdots = 6$, Euler went on to explain that by reframing the problem geometrically as quadrature, he can actually obtainActually what Euler wrote in the letter was: Terminem autem exponentis $\frac{1}{2}$ aequalis inventus est huic $\frac{1}{2}\sqrt{}(\sqrt{-1} . l\, -1)$, seu quod huic aequale est, lateri quadrati aequalis circulo, cujus diameter $= 1$. The formula essentially means $$\tfrac{1}{2}\sqrt{\sqrt{-1} \ln (-1)}$$ in modern notation. Since $\ln(-1) = i\pi + 2i\pi k, k \in \mathbb{Z}$, the expression $\tfrac{1}{2}\sqrt{\sqrt{-1} \ln (-1)}$ can be reduced to $\tfrac{1}{2}\sqrt{i(i\pi+2i\pi k)} = \frac{1}{2}\sqrt{\pi}$, when $k = -1$.: $$\tfrac{1}{2}! = \tfrac{1}{2}\sqrt{\sqrt{-1} \ln (-1)}$$ and proceeded to say that the result is equivalent to $\frac{1}{2}\sqrt{\pi} = 0.8862269$. And with this he showed that he can easily calculate $1\frac{1}{2}!$ as $$1\tfrac{1}{2}! = 1\tfrac{1}{2} \times \tfrac{1}{2}! = \tfrac{3}{4}\sqrt{\pi} = 1.3293403\ldots$$

Interestingly, another letter was addressed to Goldbach, at about the same time, by Daniel Bernoulli himself to talk about a certain cycloid problem. However, in the postscript of the letter, he briefly mentioned the use of the following infinite product to compute $x!$: $$x! = \left(A + \tfrac{1}{2}x\right)^{x-1} \frac{2}{1+x}\frac{3}{2+x}\frac{4}{3+x}\cdots\frac{A}{A-1+x}$$ It is not known which letter reached Goldbach first. It could be well that both of them reach Moscow on the same day if they are transported on the same mail carrier, even though the dates mentioned in the letter are gapped by one week.

You can roughly convince yourself that Euler's expression for $m!$ is true by the following mental process: Bring all of the factors in the denominator of the right hand side of the equation and combine it with $m!$ to form $$m!\color{brown}{(m+1)(m+2)(m+3)(m+4)\cdots}$$ which is really $1\cdot 2 \cdot 3 \cdot 4 \cdots$ since the dot-dot-dot continues indefinitely. On the other hand, the numerator can be rewritten as: $$\require{cancel}1 \cdot \bcancel{\color{brown}{2^m}} \left(\frac{2}{\bcancel{\color{brown}{2^m}}}\cdot \bcancel{\color{brown}{3^m}}\right)\left(\frac{3}{\bcancel{\color{brown}{3^m}}}\cdot \bcancel{\color{brown}{4^m}}\right)\left(\frac{4}{\bcancel{\color{brown}{4^m}}}\cdot \bcancel{\color{brown}{5^m}}\right)\cdots$$ which is apparently equivalent to $1\cdot 2 \cdot 3 \cdot 4 \cdots$ since the dots extend ad infinitum.

Sketch of the gap function $g\left(A, 1\frac{1}{2}, u\right) = \left| 1\frac{1}{2}! - \sqrt{A+\frac{3}{2u}}\prod_{k=1}^{A-1}\frac{k+1}{k+3/2}\right|$ for some values of $A$: $A = 2^3$ (in blue), $A = 2^6$ (in orange), $A = 2^9$ (in green). Apparently $u = 2$ is a special value, although the asymptotic behavior of $g$ is $g(\infty, x, u) \to 0$ for all values of $u$.

Let us now turn our attention to Bernoulli's infinite product. In modern notation, his formula can be compressed to $b(\infty, x, 2)$, where $$b(A, x, u) = \left(A+\tfrac{1}{u}x\right)^{x-1}\prod_{k=1}^{A-1}\frac{k+1}{k+x}$$ Using this, Bernoulli cited two numerical examples, one of which is $3! \approx b(16, 3, 2) = 6\frac{1}{204}$. Bernoulli's expression is correct in general since it can be transformedFirst we move all the factors in the denominator to the left and multiply them with $x!$ to form: $$x!(x+1)(x+2)\cdots(x+A-1) = (x+A-1)!$$ Then we bring $A! = 2\cdot 3\cdots A$ on the numerator to the left hand side of the equation, so that we have $${\rm LHS} = \frac{(x+A-1)!}{A!} = \frac{\bcancel{\color{white}{A!}}(A+1)(A+2)\cdots(A+x-1)}{\bcancel{\color{white}{A!}}}$$ Then we bring $(A+\frac{1}{u}x)^{x-1}$ to the denominator of LHS: $$\frac{A+1}{A+\frac{1}{u}x}\frac{A+2}{A+\frac{1}{u}x}\cdots \frac{A+x-1}{A+\frac{1}{u}x}$$ to $$\lim_{A \to \infty} \prod_{k=1}^{x-1} \frac{1+\frac{k}{A}}{1+\frac{x/u}{A}} = 1$$ Note that this relation is true for all values of $u$ so the choice of $u = 2$ in Bernoulli's infinite product is curiously interesting, for if we define $$g(A, x, u) = \left| x! - b(A, x, u)\right|$$ It is immediately clear that $u =2$ is the value associated with the best asymptotic behavior.

Sketch of the function $b\left(A, 1\frac{1}{2}, u\right) = \sqrt{A+\frac{3}{2u}}\prod_{k=1}^{A-1}\frac{k+1}{k+3/2}$, plotted from $A\in[10, 10^4]$: $u = 1$ (in blue), $u = 1.5$ (in orange), $u = 2$ (in green), $u = 2.5$ (in red), $u = 3$ (in purple).