Lets say that $\def\nn{\mathbb{N}}$$\def\rr{\mathbb{R}}$$K : \nn \to \rr$ and $\displaystyle \sum_{i=1}^\infty \frac{K(i)}{K(i+1)}$ is a convergent sum.
My conjecture is that the function $K$ must be an exponential function.
Can I have a proof of this conjecture?
EDIT: What about if $K$ does not include the factorial function?
Here is a simple proof that $K(n)$ is not only exponential but 'super' exponential in the sense that for all constants $C$, there is some $n_0$ such that $|K(n)|\geq C^n$ for all $n\gt n_0$. Let's rewrite your series as $\sum_n\frac{a_n}{a_{n+1}}$ so that we don't run out of indices; in other words, $K(n)=a_n$. (For convenience's sake I'm going to take $a_n$ positive for all $n$; this doesn't really affect the argument, but it's what requires the use of the absolute-value brackets above.)
Now, by the definition of a convergent series, the limit of the terms is 0; $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=0$. In particular, given any constant $C$, we know that there exists a $n_0$ such that $\frac{a_n}{a_{n+1}}\leq\frac1C$ for all $n\gt n_0$. But by cross-multiplication, this is equivalent to $a_{n+1}\geq Ca_n$ for all $n\gt n_0$; then by induction this implies that $a_{n_0+k}\geq C^k a_{n_0}$ for all $k$, or (shifting indices) that $a_n\geq DC^n$, where $D=C^{-n_0}a_{n_0}$. This is the result that we want 'up to a constant factor' (that factor of $D$), but that's not really relevant here - note that we could have just chosen $C$ 'twice as big' as we needed to (since this result holds for all $C$) and by the relative exponential growth of $(2C)^n$ vs $C^n$ we can eventually wipe out the constant $D$.