This is an exercise in Complex Analysis by Stein, Exercise 5.17.
Let $\{a_n\}_{k=0}^{\infty}$ be a sequence of distinct complex numbers such that $a_0 =0$ and $\lim_{k \to \infty} |a_k| = \infty$, and let \begin{align*} E(z)=z\prod_{n=1}^{\infty}{E_n (\frac{z}{a_n})} \end{align*} , where $E_k$ is Weierstrass canonical factors defined by \begin{align*} E_k (z) = (1-z) e^{z+ \frac{z^2}{2} + \cdots + \frac{z^k}{k}} \end{align*} Given complex numbers $\{b_k\}_{k=0}^{\infty}$, show that there exist integers $m_k \ge 1$ such that the series \begin{align*} F(z) = \frac{b_0}{E'(a_0 )}\frac{E(z)}{z} + \sum_{k=1}^{\infty} \frac{b_k}{E'(a_k)}\frac{E(z)}{z-a_k} \Big ( \frac{z}{a_k} \Big ) ^{m_k} \end{align*} defines an entire function that satisfies $F(a_k )=b_k $ for all $k \ge 0$.
Here is my proof, which I want to get verified. I assumed that I already know that $E(z)$ is entire, and has simple zeros only at $\{a_k\}$s.
First, let's prove that $E'(a_k) \neq 0$, so that fractions are well-defined.
Since $E$ has simple zero at $a_k$, there exists entire function $F_k$ where $E(z)=(z-a_k)F_k (z)$ and $F_k (a_k) \neq 0$. Then, \begin{align*} E'(z)=F_k (z) + (z- a_k ) {F_k}'(z) \end{align*} , thus we have $E'(a_k) \neq 0$, as desired.
Now let $m_k$ be an integer that satisfies \begin{align*} \frac{|b_k|}{|E'(a_k)|} \frac{1}{3} 4^{1-m_k} < 2^{-k} \end{align*} To prove that $F$ is entire, we only need to prove that the infinite sum converges uniformly in the closed disc $|z| \le R$.
Let $M_R$ be the supremum of $E(z)$ in $|z| \le R$. Let $N$ be a integer such that $n \ge N$ yields $|a_n| \ge \max\{RM_R , 4R\}$.
Then for $N' \ge N$, we have \begin{align*} \Big | \sum_{k=N'}^{\infty} \frac{b_k}{E'(a_k)} \frac{E(z)}{z-a_k} \Big ( \frac{z}{a_k} \Big )^{m_k} \Big | &\le \sum_{k=N'}^{\infty} \frac{|b_k|}{|E'(a_k)|} \frac{|E(z)|}{|z-a_k|} \Big | \frac{z}{a_k} \Big |^{m_k} \\ &\le \sum_{k=N'}^{\infty} \frac{|b_k|}{|E'(a_k)|} \frac{M_R}{4R-R} \frac{R}{RM_R} \Big ( \frac{R}{4R} \Big ) ^{m_k -1} \\ &= \sum_{k=N'}^{\infty} \frac{|b_k|}{|E'(a_k)|} \frac{1}{3R} \Big ( \frac{1}{4} \Big ) ^{m_k -1} \\ &\le \sum_{k=N'}^{\infty} \frac{2^{-k}}{R} = \frac{2^{1-N'}}{R} \end{align*} Since right side of the inequality converges to $0$ as $N' \to \infty$, our infinite sum converges uniformly. Thus, $F$ is entire.
Now we only need to prove that $F(a_n) = b_n$. Using that $E(a_n)=0$ and $\lim\limits_{z \to a_n} \frac{E(z)}{z-a_n} = E'(a_n)$, \begin{align*} F(a_n) = \frac{b_n}{E'(a_n)} \frac{E(z)}{z-a_n} \Big | _ {z=a_n} = b_n \end{align*}
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