Does their exist an analytic function $f(z)$ such that $f(n + \frac{1}{n}) = 0$ for all $n \in \mathbb{N}$? My approach: 
Let $a_n = n + \frac{1}{n}$. Define $f(z) = (z-a_1)\cdots(z-a_n)\cdots=\prod_{n\in\mathbb{N}}(z-a_n)$. I think that $f(z)$ satisfies the above property. But I have some reservation in defining the infinite product. So is my approach correct and can someone write down the rigorous answer to it.
 A: This should follow immediately from the Weierstrass Factorization Theorem. There are some technical details you should attend to, but your idea is almost basically correct.
https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem
A: The function you have is actually nowhere analytic and not even well defined.
Suppose that $f:\mathbb{C}\rightarrow \mathbb{C}$ where $f(x) = \lim_{n\rightarrow\infty}\prod_{i=1}^{n}(x-a_n)$ is well defined, i.e $f(x)\in \mathbb{C}$ for each $x\in \mathbb{C}$.
Pick any $x\in\mathbb{C},n\in \mathbb{N}$. 
\begin{align}
|a_n-a_m|&=|n+\frac{1}{n}-m-\frac{1}{m}|\\
&\geq |n-m| - |\frac{1}{n}-\frac{1}{m}|\\
 &\geq |n-m| - 2
\end{align}
So that $|a_n - a_m|\rightarrow\infty$ as $m\rightarrow \infty$. We also have that \begin{align}
|x-a_m|&= |(x-a_n)-(a_n-a_m)|\\
 &\geq |a_n -a_m| - |x-a_n|
\end{align}
Now, as $|a_n-a_m|\rightarrow\infty$, choose $M = 2+|x-a_n|$ so that there exists $N\in \mathbb{N}$ such that $m> N$ implies $|a_n-a_m| \geq M$ which implies $|x-a_m| \geq 2$.
Note $|x-a_i|$ is nonzero for all $1\leq i\leq N$. Thus $\prod_{1\leq i\leq N}{|x-a_i|}$ is equal to some $\varepsilon>0$. Now there exists $k\in\mathbb{N}$ such that $\frac{1}{2^k} \leq \varepsilon$. Now $\prod_{i=1}^{N+m+k}|x-a_i| \geq \varepsilon2^{m+k}\geq2^{m}$. Thus $f(x)$ must be unbounded so that $f(x)\not\in \mathbb{C}$. Hence $f$ is not well defined.
