Recently I took a complex analysis exam, and one of the problems was to prove that $$\frac{1}{\Gamma(s)} = s \prod_{n = 1}^\infty \frac{\left(1 + \frac{s}{n}\right)}{\left(1+ \frac{1}{n}\right)^s}$$ I was allowed to use the Stein-Shakarchi text, in which chapter 6, theorem 1.7 states that for all $ s \in \mathbb{C}$, $$\frac{1}{\Gamma (s)} = e^{\gamma s}s \prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right)e^{ - s/n},$$ where $\gamma$ is the Euler-Mascheroni constant. My professor took away points due to lack of rigor in one of the equalities, and said to ask other mathematicians what they thought about the proof. My proof is as follows:

We have that $$\frac{1}{\Gamma (s)} = e^{\gamma s}s \prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right)e^{ - s/n} = e^{\lim_{N \to \infty} \sum_{k = 1}^N s/k - s\log N} s \prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right) e^{- s/n}$$ The exponential terms cancel out as in the limit as $N \to \infty$ as we have that each term in the product is matched by a term from the Euler-Mascheroni term. Thus this equals $$\lim_{N \to \infty} e^{\log N^{-s}} s \prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right) = \lim_{N \to \infty} s \prod_{n = 1}^\infty \frac{1}{N^s} \left(1 + \frac{s}{n}\right)$$ $N = N/(N - 1) \cdot (N - 1)/(N - 2) \cdots 3/2 \cdot 2/1 = \prod_{n = 1}^N (1 + 1/n)$, therefore as $N \to \infty$ we have that this equals $$s \prod_{n = 1}^\infty \frac{\left(1 + \frac{s}{n}\right)}{\left(1 + \frac{1}{n}\right)^s},$$ as desired.

My professor said that this equality $$\lim_{N \to \infty} e^{\log N^{-s}} s \prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right) = \lim_{N \to \infty} s \prod_{n = 1}^\infty \frac{1}{N^s} \left(1 + \frac{s}{n}\right)$$ simply shows that $0 = 0$ and thus my argument is not rigorous. I was confused as to why this was because I only used standard rules involving limits and continuous functions, and it seems like the problems with the limit of $1/N^s$ approaching infinity are resolved in the proof of the equality that I started with (ch. 6, theorem 1/7) via the Hadamard factorization theorem. Is this rigorous or not? If so, what is my mistake? Thank you very much.

  • 3
    $\begingroup$ It seems to me that there are at least two places where limits are exchanged and are not justified: when you cancel the exponential factors, and when you replace the limit for $N^{-s}$ with the term in the inside the infinite product. $\endgroup$
    – shoteyes
    Jun 26, 2021 at 2:09
  • 1
    $\begingroup$ I just edited it, thanks for the heads up $\endgroup$
    – Rough L
    Jun 26, 2021 at 2:21
  • 1
    $\begingroup$ As written, the step $$\lim_{N \to \infty} e^{\log N^{-s}} s \prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right) = \lim_{N \to \infty} s \prod_{n = 1}^\infty \frac{1}{N^s} \left(1 + \frac{s}{n}\right)$$ just doesn't make sense at all (ignoring issues about dealing with limits correctly). On the left side you have one factor of $N^{-s}$ but on the right you have infinitely many. $\endgroup$ Jun 26, 2021 at 3:43
  • 1
    $\begingroup$ I think you mean for the factor $\frac{1}{N^s}$ to be outside the product on the right side there? $\endgroup$ Jun 26, 2021 at 3:45
  • $\begingroup$ The product $$\prod_{n=1}^\infty\left(1+\frac sn\right)$$ diverges. $\endgroup$
    – robjohn
    Jun 26, 2021 at 10:42

1 Answer 1


You are repeatedly making assertions of the form $$\left(\lim_{N\to\infty}a_N\right)\cdot\left(\lim_{N\to\infty}b_N\right)=\lim_{N\to\infty}\left(a_Nb_N\right),$$ but this is only valid assuming that you know both limits on the left side exist. For instance, when you go from $$\left(\lim_{N \to \infty} N^{-s}\right)\prod_{n = 1}^\infty \left(1 + \frac{s}{n}\right)$$ to $$\prod_{n = 1}^\infty \frac{\left(1 + \frac{s}{n}\right)}{\left(1 + \frac{1}{n}\right)^s}$$ you are applying this with $$a_N=N^{-s}=\prod_{n = 1}^N (1 + 1/n)^{-s}$$ and $$b_N=\prod_{n = 1}^N \left(1 + \frac{s}{n}\right).$$ This is valid only if you already know that $(a_N)$ and $(b_N)$ converge. But they do not converge, at least for most values of $s$. (In fact, if $s>0$ then $(a_N)$ converges to $0$, which I think is what your professor was alluding to and should immediately set off alarm bells--it means that your expression (if it is well-defined) can only be equal to $0$, since it is a product of factors, one of which is $0$!)

There is a similar issue when you originally cancelled out the exponential factors. That involves separating $$e^{\lim_{N \to \infty} \sum_{k = 1}^N s/k - s\log N}$$ as $$e^{\lim_{N\to\infty}- s\log N}\cdot e^{\lim_{N \to \infty} \sum_{k = 1}^N s/k}$$ (and then recombining the latter factor with your other infinite product), but this only makes sense if you know that the two limits on the second line actually exist. In fact, they do not.

Essentially, what you have done is taken a non-absolutely convergent series, split it as a sum of a bunch of divergent series, then grouped together the terms of the divergent series so that they cancel out and give something convergent again. (Except you did it with a product instead of a series; if you take logs of everything then that's what you would have done to the corresponding series.) There is no reason to expect the series to have the same sum after such a rearrangement.

  • $\begingroup$ I see where the problem is, at least regarding to the $N^{-s}$ factor. Seems I'm too used to working with sums. With convergence and cancellation in mind, how would I rigorously prove this? Thank you for the explanation $\endgroup$
    – Rough L
    Jun 26, 2021 at 8:29
  • 1
    $\begingroup$ I don't know off the top of my head. I would guess that maybe you can try applying the Hadamard factorization theorem to the function $s \prod_{n = 1}^\infty \frac{\left(1 + \frac{s}{n}\right)}{\left(1+ \frac{1}{n}\right)^s}$ similar to how Theorem 1.7 applies it to $1/\Gamma$. $\endgroup$ Jun 26, 2021 at 16:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .