It is well known that Euler have proved that $e=[2, 1, 4, 1, 1, 6, \ldots, 2n, 1,1, \ldots]$ and that $e$ is a transcendental number by Hermite's evidence.
Let us consider the function $e(T)$ presented by the following continued fraction: $$e(T)=[2T, T, 4T, T, T, 6T, ....,2nT,...]\in\mathbb{Q}((T^{-1})).$$ Is it transcendental?