Suppose $\langle z_n \rangle_{n\ge0}$ is a sequence of complex numbers such that $\sum_{n=0}^\infty z_n$ converges,Given that $f$ is an entire function such that $f(z_n)=n$, then
$f\equiv 0$.
$f$ is unbounded.
No such $f$ exists.
$f$ has no zeroes.
Suppose $\langle z_n \rangle_{n\ge0}$ is a sequence of complex numbers such that $\sum_{n=0}^\infty z_n$ converges,Given that $f$ is an entire function such that $f(z_n)=n$, then
$f\equiv 0$.
$f$ is unbounded.
No such $f$ exists.
$f$ has no zeroes.
As Gerry pointed out, $z_n \rightarrow 0$. Entire functions are continuous, so think about what we can say about $f(0)$.
Also, what you have written about $g(z)$ doesn't make any sense. You can't say $g(z) = f(z) - n$ has zeros at all of the $z_k$ because for the definition of $g(z)$ to make sense, $n$ must be fixed. So it would have a zero at $z_n$ but not at $z_k$ if $k \neq n$.