Show that This infinite product is entire

Question

This question is from Ponnusamy and silvermann complex analysis and I am making some mistake in this question and unable to find it.

Show that $\prod_{n=1}^{\infty} (1-z/n) e^{z/n +5z^2/n^2}$ is entire.

$1+ f_n (z)= (1-z/n) e^{z/n +5z^2/n^2}$ and finding a upper bound on $f_n (z)$ . $1+ f_n (z) = exp( Log(1-z/n) + z/n+ 5 z^2 /n^2)$ For $|z|\leq R$ then choose N large enough such that |z/n|<1 for all $n \geq N$ and using the identity$ Log(1-z/n) = -\sum_{k=1}^{\infty} 1/k (z/n)^k$ and the fact that|R/n|<1.

So, $|Log (-\sum_{k=1}^{\infty} 1/k (z/n)^k +z/n +5 z^2/n^2) | \leq -5 -1/2 R^2/n^2 $, where I used Formula of infinite geometric series.

Now $|f_n(z)| \leq exp|e^{-5 -1/2 R^2 /n^2 -1}-1| exp( -5 -1/2 R^2 /n^2)$ -1 and using $e^x -1 \leq x e^x for all x \geq 0 $ we get $|f_n(z)|\leq 5+ 1/2 R^2/ n^2 $ and if I sum $|f_n (z)| $ from N to $\infty$ i get Infinity as 5 is also there .

So, I think while approximating I am doing some mistake and I am unable to find it. Can you please point that out.

If there is some other method of proving it entire that is also welcome.

Thanks!

Answer 1

Because $f(z)=\frac{\log(1-z)+z}{z^2}$ is analytic thus continuous for $|z|< 1$ then $$C= \sup_{|z|\le 1/2}|f(z)|<\infty$$ gives $|\log(1-z/n)+z/n|\le Cz^2/n^2$ for $|z|<n/2$.

Answer 2

Hint.

Theorem: Suppose $C$ is an open subset of $\Bbb C$ and $(f_n)_n$ is a sequence of complex functions that are analytic on $C,$ such that $f_n$ converges uniformly on $D$ to $f$ whenever $D$ is a compact subset of $C$. Then $f$ is analytic on $C.$