How to know if some series doesn't converge to a rational function I was looking into a previous exam from 2011 of a course I am taking of Complex Analysis, and they ask

Which of the following series converge to a rational function in some domain?
  $$\sum_{k=0}^\infty \frac{1}{k!+k^2+k}z^{k^2+2k}\\\sum_{k=0}^\infty \frac{2^k}{(1+z^2)^k}\\\sum_{k=0}^\infty\frac{1}{k! (z-k)^k}$$

I had no problem with the second and third ones. The second one converges to $\frac{1}{1-\frac{2}{1+z^2}}$ given that $\left|\frac{2}{1+z^2}\right|<1$, and the third one has an infinite number of singularities so it can't be a rational function.
I don't know how to verify that in the first series. Is there any general method to see if some power series converge to a rational function, or in this particular case, a way to see if this series does?
Edit: Now I am not that convinced about my argument for the third series (Since the series might only converge in a bounded domain from which the number of singularities would be finite). Is there anything wrong with it or any way to formalize it further? I also just found the same argument here.

The first series converges uniformly to some $f$ in $\overline {\mathbb D},$ and diverges for all other $z.$ Thus $f$ is analytic in $\mathbb D.$ Suppose $f=R$ in some domain $U,$ where $R$ is a rational function. Then $U\subset \mathbb D.$ Let $P$ be the set of poles of $R.$ Then $f,R$ are both analytic in $\mathbb D\setminus P,$ which is also a domain. By the identity principle, $f=R$ in $\mathbb D\setminus P.$ Since $f$ has no singularities in $D(0,1),$ neither does $R.$ Therefore $f=R$ in $\mathbb D.$ Now $f$ extends continuously to $\overline {\mathbb D},$ hence so does $R.$ Therefore $R$ has no poles in $\overline {\mathbb D}.$ It follows that $R$ is analytic in some $D(0,r), r>1.$ Thus $f$ has an analytic extension to $D(0,r).$ Therefore the power series defining $f$ converges in $D(0,r),$ contradiction.
Added later For the third series: Let $\Omega =\mathbb C\setminus \{0,1,2,\dots\}.$ Note that $\Omega$ is an open connected set. The given series converges uniformly on compact subsets of $\Omega$ to an analytic function $f,$ and $f$ has a pole at each point of $\{0,1,2,\dots\}.$ (Please ask if you have questions on this.) Let $R$ be a rational function, and let $P$ be the set of poles of $R.$ Then both $f,R$ are analytic on $\Omega \setminus P,$ which is also an open connected set. Suppose $f=R$ on an open subset of $\Omega \setminus P.$ Then by the identity principle, $f=R$ everywhere in $\Omega \setminus P.$ It follows that $R$ has a pole at each point of $\{0,1,2,\dots\}.$ This is a contradiction, since $R$ has at most a finite number of poles.
