non-constant entire function $f$ such that $f(n+\dfrac{1}{n})=0\forall n\in \Bbb N$?

Does there exist a non-constant entire function $f : \mathbb{C}\to\mathbb{C}$ such that $f(n+\dfrac{1}{n})=0$ for all $n\in \Bbb N$?

Let $f$ be a non-constant entire function such that $f(n+\dfrac{1}{n})=0\forall n\in \Bbb N$.

Then $f(2)=0;f(3+\frac{1}{3})=0$ and so on.But the problem is the set of zeros of $f$ does not have a limit point.

• The Weierstrass factorization theorem allow you to find an entire function such that has zeros exactly at a sequence $\{z_n\}_{n\geq 1}$ such that $z_n \to \infty$. Oct 14 '16 at 5:12
• $\prod_{n=1}^\infty (1-\frac{z}{n+1/n})e^{z/(n+1/n)}$ Oct 14 '16 at 5:13
• Otherwise, you should be able to modify $\displaystyle\frac{1}{\Gamma(\frac{z+(z^2-4)^{1/2}}{-2})}$ such that it is entire Oct 14 '16 at 5:15
• Oct 14 '16 at 5:20

$$f(z) = \prod_{n =1}^\infty \left(1-\frac{z^2}{(n+1/n)^2}\right)$$
determines a nonconstant entire function of $z$ with zeroes at $z= \pm (n+1/n)$.
• @Bruno Joyal...What will be the infinite product if $f(a+1/n)=0$, $a\in\mathbb R$? Oct 14 '16 at 14:13