$\prod_{n=1}^{\infty} (1- z/ a_n) $ is entire iff $\sum_{n=1}^{\infty} 1/(z-a_n) $ is meromorphic

Question

This question was asked in a masters exam previous year paper and I was unable to prove it.

Show that the $$\prod_{n=1}^{\infty} (1- \frac{z}{a_{n}}) $$ is entire iff $$\sum_{n=1}^{\infty} \frac{1}{z-a_n} $$ is meromorphic.

$\prod_{n=1}^{\infty}( 1- \frac{z}{a_{n}})$ is entire if Convergence of infinite product is uniform and that happens if: $$\sum_{n=1}^{\infty} \frac{z}{a_{n}}$$ converges for all z. But I am not able to correlate it with: $$\sum_{n=1}^{\infty} \frac{1}{z-a_n} $$.

Can you please help by telling which result should be helpful.

Thank you!

Answer 1

You probably meant if $a_n\ne 0,f_N(z)=\sum_{n\le N} \frac1{z-a_n}$ and $\lim_{N\to \infty}f_N(z)$ converges locally uniformly away from the $a_n$ to a meromorphic function, then $\prod_n (1-z/a_n)$ is entire.

In which case it is obvious:

$\lim_{N\to \infty}\exp(\int_0^z f_N(s)ds)$ converges locally uniformly to an analytic function away from the $a_n$.

Around the $a_n$ we use the maximum modulus principle, uniform convergence on $|z-a_n|=r$ implies uniform convergence on $|z-a_n|\le r$.

Answer 2

@dezdichado gives the only if direction. If $\prod_{n=1}^\infty \left(1-{z\over a_n}\right) = :f$ is entire, then taking logarithmic derivative we get $${f'(z)\over f(z)} = \sum_{n=1}^\infty{-{1\over a_n}\over \left(1-{z\over a_n}\right)} = \sum_{n=1}^\infty {1\over z-a_n}.$$ Since the LHS is meromorphic, done.

Now suppose $\sum_{n=1}^\infty{1\over z-a_n}$ is meromorphic. Then it converges absolutely and uniformly on each compact subset except poles. In particular, setting $z =0$, we get $$\sum_{n=1}^\infty {R\over|a_n|}<\infty,\quad\forall R>0.$$ In particular, the infinite product $$\prod_{n=1}^\infty\left(1-{z\over a_n}\right)$$ is entire.