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In Euler's first (known) letter to Goldbach on October 13, 1729 he mentions an infinite product that interpolates the factorial function:

$$s!=\lim_{n\rightarrow\infty}\frac{n!}{\left(s+1\right)\left(s+2\right)\cdots\left(s+n\right)}\left(n+1\right)^{s}$$

for any integer $s$. However this limit exist for all complex $s$ except for the negative integers.

My question is: How did Euler deduce this infinite product?

I've seen how Gauss deduced a very similar formula:

$$\Pi\left(s\right)=\lim_{n\rightarrow\infty}\frac{n!}{\left(s+1\right)\left(s+2\right)\cdots\left(s+n\right)}n^{s}$$

however the method Gauss used doesn't explain how the term $(n+1)^s$ appears in Euler's representation.

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  • $\begingroup$ I think your claim is not true. See my answer. $\endgroup$
    – glebovg
    Jun 21, 2014 at 23:01

2 Answers 2

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Have a look at this article. In particular, p. 853.

Also, I think your claim is not true because $$\lim_{n \to \infty} \frac{n!}{\prod_{k = 1}^n (s + k)} \cdot (n + 1)^s = \lim_{n \to \infty} \frac{n!}{\prod_{k = 1}^n (s + k)} \cdot n^s$$ since $(n + 1) \sim n$, that is, $(n + 1)$ is asymptotically equal to $n$. In other words, $\lim_{n \to \infty} (n + 1)/n = 1$, or, in the 18th century jargon, if $n$ is an infinitely large number, then $n = n + 1$.

However, $\Gamma(s) = (s - 1)!$ provided $s \in \mathbb{N}$, so the result you attributed to Gauss would imply that $\Gamma(s) = s!$, which is false for all $s \ne 1$. Gauss never used Legendre's notation, namely, $\Gamma(s) = \int_0^\infty x^{s - 1} e^{-x} dx$, instead he introduced his eponymous function $\Pi(s)$, which equals $\Gamma(s + 1)$, and $\Pi(s) = s!$ whenever $s \in \mathbb N \cup \{0\}$. I think you meant $$\Pi(s) = \lim_{n \to \infty} \frac{n!}{\prod_{k = 1}^n (s + k)} \cdot n^s.$$

Edit: By the way, here a "modern" proof I know of:

Note that $$\Gamma(s) = \int_0^\infty t^{s - 1} \lim_{n \to \infty} \left(1 - \frac{t}{n}\right)^n \, dt = \lim_{n \to \infty} \int_0^n t^{s - 1} \left(1 - \frac{t}{n}\right)^n \, dt.$$ (This is left as an exercise.) Repeated integration by parts with $u = (1 - t/n)^n$ and $v' = t^{s - 1}$ gives $$\Gamma(s) = \lim_{n \to \infty} \left(\frac{1}{s} \cdot \frac{n - 1}{(s + 1)n} \cdot \frac{n - 2}{(s + 2)n} \cdots \frac{1}{(s + n - 1)n} \int_0^n t^{s + n - 1} \, dt\right).$$ Therefore, $$\Gamma(s) = \lim_{n \to \infty} \frac{n!}{n^n} \prod_{k = 0}^n \frac{n^{s + n}}{s + k} = \lim_{n \to \infty} \frac{n^s}{s} \prod_{k = 1}^n \frac{k}{s + k},$$ which is equivalent to the formula you seek since $s\Gamma(s) = \Gamma(s + 1) = s!$ if $s \in \mathbb{N} \cup \{0\}$.

However, it is also Euler's product [p. 853] in disguise for if $s \in \mathbb{N} \cup \{0\}$ then $s\Gamma(s) = \Gamma(s + 1) = s!$ and \begin{align*} s! &= \lim_{n \to \infty} n^s \prod_{k = 1}^n \frac{k}{s + k} \\ &= \lim_{n \to \infty} (n + 1)^s \prod_{k = 1}^n \frac{k}{s + k} \\ &= \lim_{n \to \infty} \frac{(n + 1)^s n!}{(s + 1)(s + 2)(s + 3) \cdots (s + n)} \\ &= \left(\frac{2}{1}\right)^s \frac{1}{s + 1} \times \left(\frac{3}{2}\right)^s \frac{2}{s + 2} \times \left(\frac{4}{3}\right)^s \frac{3}{s + 3} \times \cdots \end{align*}

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  • $\begingroup$ You are completely correct, I've already modified the question. The equality of the limits has given me two ways to deduce Euler's infinite product: Gauss' way and Euler's which is explained in both the article you shared and the previous answer. $\endgroup$ Jun 23, 2014 at 1:41
  • $\begingroup$ @AlejandroDeLasPeñas Great. I also added a proof I know of. I hope it helps. $\endgroup$
    – glebovg
    Jun 23, 2014 at 5:48
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Euler, of course, was prolific in his investigation of infinite series and products.

Supposedly, he examined an infinite product with a salient pattern

$$\left(\frac{2^n}{1^n}\frac{1}{n+1}\right)\left(\frac{3^n}{2^n}\frac{2}{n+2}\right)\left(\frac{4^n}{3^n}\frac{3}{n+3}\right)\ldots \ (*)$$

and observed that if $n$ is a positive integer, then the product converges to $n!$.

Using your notation ($s \in \mathbb{N})$, the partial product for $(*)$ can be expressed in the form where $(n+1)^s$ appears

$$P_n(s)=\frac{n!}{(s+1)(s+2)\ldots(s+n)}(n+1)^{s}.$$

After some cancelation and manipulation we get for $n >s,$

$$P_n(s)=s!\left(\frac{1+1/n}{1+2/n}\right)\left(\frac{1+1/n}{1+3/n}\right)\ldots\left(\frac{1+1/n}{1+s/n}\right),$$

and it is apparent that $P_n(s) \rightarrow s!$ as $n\rightarrow \infty.$

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  • $\begingroup$ Sorry, but your $(*)$ doesn't look converging to me, can you explain the mathematical manipulation you did? $\endgroup$
    – user427802
    Aug 17, 2019 at 18:46
  • $\begingroup$ @AbhasKumarSinha: The reference to (*) is just part of the story about Euler. To see the convergence you have to define carefully what a product followed by $\ldots$ means. Euler was not always careful in this regard. $\endgroup$
    – RRL
    Aug 17, 2019 at 19:35
  • $\begingroup$ Take $n$ fixed and $m > n$. The expression (*) means the limit of partial products $P_m$ as $m \to \infty$. In this case, $P_m = \frac{2^n}{1^n}\frac{3^n}{2^n} \cdots \frac{(m+1)^n}{m^n}\frac{m!}{(n+1)(n+2) \cdots (n+m)}$ which I showed converges to $n!$ as $ m \to \infty$ below (replacing $m$ with $n$ and $n$ with $s$. $\endgroup$
    – RRL
    Aug 17, 2019 at 19:40

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