A nonzero sequence $a(n)$ is called hypergeometric iff $a(n+1)/a(n) =$ $p(n)/q(n)$ for almost all $n \in \mathbb{N}$ and for polynomials $p,q$.
In the book Hypergeometric Summation by Wolfram Koepf it is mentioned that $2^n+1$ is not hypergeometric even though $2^n$ and $1$ are. It is clear that those are indeed hypergeometric. I wanted to show that $2^n+1$ is not hypergeometric but unfortunately I have no idea how. I could think about it in an analytical way (e.g. limits) to get restraints on how the polynomials would have to look like but this is not really helping me.
Is there a good general way to proof that sequences are not hypergeometric? How could I go about this example in particular?
For instance I managed to show that $\sqrt{n}$ is not hypergeometric by squaring the ratio and thinking about divisibility of the resulting polynomials. However, I could not adapt the proof for this example as I do not not how to get polynomials such that I can use divisibility properties.
Any help is appreciated!
Nonetheless, limits are the right way to go. If you have $a(n)=2^n+1$, that implies $a(n+1)/a(n)\to2$ as $n\to\infty$. Of course, that would be compatible with $a(n+1)/a(n)=p(n)/q(n)$, as long as the degree of the polynomial $2\,q(n)-p(n)$ is smaller than the degree of $q(n)$.
However, if the degree of $q(n)$ is $k$, and the degree of $2\,q(n)-p(n)$ is $l<k$, we would have $$\lim_{n\to\infty}n^{k-l}\left(2-\frac{p(n)}{q(n)}\right)\neq0.$$ But $$2-\frac{a(n+1)}{a(n)}=\frac1{2^n+1},$$ and $$\lim_{n\to\infty}\frac{n^{k-l}}{2^n+1}=0$$ for any $k,l$