As$\ n \to \infty$, can a transcendental function$\ f\left(1+ \frac{1}{n}\right)$ to the power of$\ n$ tend to a rational power of$\ e$?

Question

Let$\ f(n)$ be a transcendental function$\ \ne e^{g(n)}$, for any function$\ g(n)$. Does$$\ \lim_{n \to \infty} \left(f\left(1+ \frac{1}{n}\right)\right)^n =e^{ -k} = \lim_{n \to \infty} \left(1 - \frac{k}{n}\right)^n $$ imply the irrationality of$\ k$? I have no means to think whether this has a positive or negative answer, but I do think the number$\ k$ at issue (which obviously has other properties) is irrational, hence the question.

Answer 1

Example.
$f(x) = 1/\Gamma(x)$ is certainly a transcendental function. It has zeros, so it is not exp of an entire function. But your value $k$ is Euler's constant $\gamma$. Which has not yet been proved irrational...

more
If $c$ is any nonzero constant, then $\sin(cx)$ is a transcendental function. But your value $k$ is then $c \cot c$ if my computations are right. And certainly there is some constant $c \approx -4.493$ so that $c \cot c = 1$.

Answer 2

$$f(x) = {e}^{-k(x-1)}$$

This function should be transcendental and satisfy the conditions you've shown, but $k$ can be rational.

Answer 3

Well if $f(1) = 1$ and $f$ is differentiable at $1$, then $\lim (f(1+1/n))^n = \exp(f'(1))$.

Now, there is absolutely no relationship whatsoever between differentiable functions whose derivative at $1$ is rational and "transcendental functions" (whatever that means), or wether $f$ is of the form $\exp(g)$ (or equivalently, functions with no zeros).