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I want to prove that the product $$\prod_{n=1}^\infty \left(1+\frac{z}{n} \right) \mathrm{e}^{-\frac{z}{n}}$$ converges absolutely, and uniformly on compact subsets of $\mathbb C$:

My book (Ahlfors) defines the absolute convergence of the product $\prod a_n$ by the absolute convergence of the series $\sum \text{Log } a_n$, where only a finite number of terms can be zero, and they must be removed from the sequence. ($\text{Log}$ denotes the principal branch of the logarithm, where the argument is in $(-\pi ,\pi]$.)

Thus, in order to prove absolute convergence, I fix $z \in \mathbb C$, and look at the series whose terms are $\text{Log } \left[ \left(1+\frac{z}{n} \right) \mathrm{e}^{-\frac{z}{n}} \right]$. It is well known that if $\text{Arg }z_1,\text{Arg }z_2$ both lie in $(-\frac{\pi}{2},\frac{\pi}{2})$, then $\text{Log } (z_1z_2)= \text{Log } z_1+\text{Log } z_2$. For large enough $n$ both factors $z_1=\left( 1+\frac{z}{n} \right),z_2=\mathrm{e}^{-\frac{z}{n}}$ satisfy this, so $\text{Log } \left[ \left(1+\frac{z}{n} \right) \mathrm{e}^{-\frac{z}{n}} \right]=\text{Log } \left(1+\frac{z}{n}\right)+\text{Log } \mathrm{e}^{-\frac{z}{n}}$. It is also possible to prove that $\text{Log } \mathrm{e}^{-\frac{z}{n}}=-\frac{z}{n}$ for large enough $n$. The problem thus reduces into the convergence of $$\sum \left\lvert \text{Log } \left(1+\frac{z}{n} \right)-\frac{z}{n} \right\rvert $$

Taylor's theorem states that $$\text{Log }(1+w)=w+w^2 g(z) .$$ I've estimated the remainder and found that for $|w|<\frac{1}{2}$, $$| g(w)| \leq 2(\ln 2+\pi), $$ therefore, for large enough $n$ (so that $|\frac{z}{n}| < \frac{1}{2}$), the terms of the last series are dominated by $2(\ln2+\pi)\frac{|z|^2}{n^2}$, and $\sum 2(\ln2+\pi)\frac{|z|^2}{n^2}$ converges.

I deliberately left the $\Sigma$'s unindexed, since $z$ might be a zero of one of the factors. In that case only a tail of the series is taken into account.

The proof of uniform convergence on compact sets follows the same lines. Let $K \subset \mathbb C $ be a compact set, as such it lies in some ball $|z| \leq M$. Taking large enough $n$, we find similarly, that the series $\sum \left\lvert \text{Log } \left[ \left(1+\frac{z}{n} \right) \mathrm{e}^{-\frac{z}{n}} \right] \right\rvert$ is dominated by $\sum 2(\ln2+\pi)\frac{M^2}{n^2}$ which converges. The Weierstrass M-test says that the series is uniformly convergent, and hence, so is the product.

I'd love to hear any thoughts/remarks about these proofs.

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I find it OK. The general case is considered in W. K. Hayman, Meromorphic functions, Claredon Press, Oxford, 1964, paragraph 1.10 and here, paragraphs 2.3, 2.4. – user64494 Aug 8 '13 at 18:48
up vote 1 down vote accepted

Looks fine. It's natural to take logarithm for a product, but I think there should be a more direct way to define absolute convergence of a product besides literally defining it in terms of the sum of logarithm.

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