Critique of this proof of $\frac{1}{\Gamma(s)} = s \prod_{n = 1}^\infty \frac{\left(1 + \frac{s}{n}\right)}{\left(1+ \frac{1}{n}\right)^s}$

Question

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.

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.