$\frac{f'(z)}{f(z)}= \sum_{n=1}^{+ \infty}\frac{f'_{n}(z)}{f_{n}(z)}$

Question

I 've found this exercise.

Let $\{f_{n}\}$ be a sequence of holomorphic functions on a given domain $\Omega$. Suppose that $\prod_{1}^{\infty}f_{n}$ converges uniformly on compact subsets of $\Omega$ to $f$.

a) Show that $$\sum_{k=1}^{\infty}(f'_{k}(z)\prod_{n \neq k}f_{n}(z))$$ converges uniformly on compact subsets of $\Omega$ to $f'$.

b) Suppose that $f$ is non-zero on a given compact set $K \subset \Omega$. Show that $$\frac{f'(z)}{f(z)}= \sum_{n=1}^{+ \infty}\frac{f'_{n}(z)}{f_{n}(z)}$$ and that the convergence is uniform on K.

Any hint ? In particular: is admissible to " divide " a series by an infinite product an to make some semplifications ?

Answer 1

Hints: a) For $n \in \mathbb{N}$, let

$$\begin{align} p_n(z) &= \prod_{k=1}^n f_k(z),\\ q_n(z) &= \prod_{k=n+1}^\infty f_k(z). \end{align}$$

By assumption, $(p_n)$ converges compactly to $f$. Then consider

$$p_n'(z)\cdot q_n(z).$$

b) Since $1/f(z)$ is bounded, that follows from a).

In particular: is admissible to " divide " a series by an infinite product an to make some semplifications ?

Yes, the point of the exercise is to prove that you can do that (if the convergence of both is locally uniform).