Prove that a bounded analytic function in the right half-plane which vanishes at each positive integer is identically zero. I want to solve the example below, but I can not.
Prove that a bounded analytic function in the right half-plane which vanishes at each positive integer is identically zero. 
Please, you will be very grateful if someone would help resolve the communities in this example. Previously thank you.
 A: $\textbf{Claim:}$ $f$ analytic and bounded in the half plane: $Re(z)>0$ having zeros at
$z=1,2,3,...$ implies that $f$ is identically zero.
Consider $g(z)=f(\frac{1}{1+z}-\frac{1}{2})$ for $|z|<1$
(note: $\frac{1}{1+z}-\frac{1}{2}$ maps $|z|<1$ onto the right half plane ... it is nothing but
 a Moebius transformation)
$g$ is analytic in the (open) unit disk and bounded there and $g(z)=0$ whenever $\frac{1}{1+z}-\frac{1}{2}=n$ for $n$ a positive integer, or:
$$z=-1+\frac{1}{n+\frac{1}{2}}$$
Let me call 
$$-1+\frac{1}{n+\frac{1}{2}}=z_n$$
 and note that the product $\prod(|z_n|)$ diverges to zero!
We want to show that $g(z)$ is identically zero. As it is analytic at
zero, we will show that it has a zero of "infinite mutliplicity" there
( if $g$ is analytic at zero and is not identically zero, it has a
zero of some finite order ... that is what we will contradict).
Let $B_k(z)=\prod_{n=1}^{k}(1-z\cdot z_n')/(z-z_n)$ where $'=$conjugate (of
course, $z_n$ is real, so I don't really need the prime). 
$g(z)\cdot B_k(z)$ is analytic (as $g$ has zeros at $z_n$) and by the maximum modulus principal we
have:
$$|g(z)\cdot B_k(z)|<=\textrm{max_value_of_this_for_(|z|=r)}$$
whenever $|z|<r$.
Take the $\sup$ of the right hand side as $r\rightarrow 1$. $|B_k(z)|$ has a limit of $1$
as $|z|\rightarrow1$ and $|g|$ is bounded by, let's say, M (the original bound we
have for the original function on the half plane).
We have $|g(z)|\leq\frac{M}{|B_k(z)|}$ for all $|z|<1$ and all $k$.
Put in $z=0$:
$|g(z)|<=M\cdot\prod_{n\leq k}|z_n|$ and take the limit as $k\rightarrow\infty$.
The product diverges to zero, so $g(0)=0$.
Well, then, $\frac{g(z)}{z}$ is analytic so do this again, replacing $g$ by $\frac{g}{z}$
$|\frac{g(z)}{z}\cdot B_k(z)|\leq M $ (since we use maximum modulus and as $|z|\rightarrow 1$, both
$|B_k(z)|$ AND $|z|$ tend to one). Again, since the product of $|z_n|$
diverges to zero, $\frac{g(z)}{z}$ has a "value" (value of $h(z)=\frac{g(z)}{z}$, the
analytic function with the first zero at $z=0$ removed) of $0 $... or $g$ has
a double zero at $z=0$.
... a triple zero! Etc. etc.
(if $g$ is not identically zero, there exists $h(z)$ with $h(0)<>0$,
$g(z)=z^a\cdot h(z)$ for some positive integer $a$ ... replace $g$ by $h$ in the
argument above to show that $h(0)=0$, a contradiction).
