Uniform convergence of product in ball $|z|<R$

Question

Find a polynomial $P(z)$ such that $$\prod_{n=1}^\infty \left(1+\dfrac{z}{\sqrt{n}}\right)e^{-P(z)}$$ converges uniformly in every ball $|z|<R$.

Maybe we should take log to get that the sum $$\sum_{n=1}^\infty \left(\log\left(1+\dfrac{z}{\sqrt{n}}\right)-P(z)\right)$$ converges.

By Taylor series, we can write $\log\left(1+z\right)=z+g(z)z^2$ for $|z|<1$, where $g(z)$ is analytic in $|z|<1$.

Let $B = \sup_{|z|<1/2}|g(z)|$.

Now, for $|z|<R$ and $n>4R^2$, so that $g\left(\dfrac{z}{\sqrt{n}}\right)<B$, we can write $$\left|\log\left(1+\dfrac{z}{\sqrt{n}}\right)-P(z)\right|=\left|\dfrac{z}{\sqrt{n}}+g\left(\dfrac{z}{\sqrt{n}}\right)\cdot\left(\dfrac{z}{\sqrt{n}}\right)^2-P(z)\right|$$

So we should at least select $P(z)$ to contain the term $\dfrac{z}{\sqrt{n}}$. What to do next?

Answer 1

Am I missing something? An infinite product of nonzero complex numbers $\Pi_{n=1}^{+\infty} a_n$ has no chance of converging if you do not have $\lim_{n\rightarrow +\infty} a_n = 1$.

Here the limit is $e^{-P(z)}$, so your only try is $P(z) =0$. But it fails.