Why is $e$ transcendental, even though it can be written as a sum of algebraic numbers? I am, of course, referring to the Maclaurin series for $e$ evaluated at $1$. Then we have $$e = \sum_{n = 0}^{\infty}\frac{1}{n!} = 1+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}...$$Then $e$ is the sum of an infinite series of algebraic numbers, which would imply $e$ is algebraic. What am I missing? Is it the fact that the sum is infinite? I know $e$ is transcendental because it cannot be expressed by a polynomial with integer coefficients, so why is it that it can be expressed by a sum of algebraic numbers?
 A: What, essentially, you are wondering is why limits do not preserve algebraicity. Specifically, you have noted that even though the partial sums $$S_n=\sum_{k=0}^{n}\frac1{k!}$$
are all rational, their limit $\lim_{n\to\infty}S_n=e$ is not.
And frankly, this is the case because of the way we have defined real numbers. That is, any $x\in\Bbb R$ can be written as a decimal $$x=p.q_1q_2q_3...$$
where $p\in\Bbb Z$ and $q_k\in\{0,1,...,9\}$. Hence every number $$a_M=p.q_1...q_M=10^{-M}\lfloor 10^M x\rfloor$$
is rational and the sequence $(a_M)_{M\ge0}$ converges to $x$. In other words, every real number, regardless of whether or not it's rational, has a sequence of rationals which converges to it.
So that, I guess, means that being the limit of a rational sequence isn't a very interesting property, unless you're into that kind of thing.
At the end of the day, it turns out that transcendence and being the limit of a rational sequence have actually nothing to do with each other, even though it may seem like they might at the surface level. For a really nice transcendence proof of $e$ (and $\pi$), check out this video by the wonderful mathematics youtuber Mathologer.
