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?
Why is $e$ transcendental, even though it can be written as a sum of algebraic numbers
transcendental-numbers
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Short answer: yes, there are numbers worse than $e$ and $\pi$. Almost all real numbers are worse. There are uncountably many reals, whereas any class of "nice transcendental numbers" you can think of will be only countable (and I propose exponential periods as a good class of numbers of the kind you describe).
The class of numbers which are roots of polynomials with rational coefficients are the algebraic numbers. The numbers which are definite integrals of algebraic functions with algebraic bounds are called the ring of periods. It is a larger class of numbers, including many familiar (suspected) transcendental numbers such as $\pi$, $\log 2$, $\zeta(3)$, and $\Gamma(p/q)^q$. See this nice primer by Kontsevich and Zagier.
Actually some common numbers like Euler's number $e$, the Euler-Mascheroni constant $\gamma$, and $1/\pi$ are still (suspected to be) missing in the ring of periods. So Kontsevich and Zagier go further and extend the ring to exponential periods, defined as integrals of products of exponentials of algebraic functions with algebraic functions. This gives a class of numbers that include "all algebraic powers of $e$, values of $\Gamma$ at rational arguments, values of Bessel functions, etc."
I assume that the exponential periods still don't include $\gamma$, because finally they claim that if you extend the class further by adding it, then you have "all classical constants in an appropriate sense" (whatever that means).
In this primer on exponential motives by Fresán and Jossen, according to Belkane and Brosnan, $\gamma$ is an exponential period, as witnessed by the integrals $\gamma=-\int_0^\infty\int_0^1e^x\frac{x-1}{(x-1)y+1}\,dy\,dx$ or $\gamma=-\int_0^\infty\int_1^x\frac{1}{y}e^{-x}\,dy\,dx$.
Anyway, the rational numbers are countable. The algebraic numbers are countable. The ring of periods is countable. The exponential periods are countable. And including $\gamma$ certainly still leaves you with a countable class of periods.
But the real numbers are uncountable, so it remains the case that most real numbers are not periods, not exponential periods, and cannot be written in any of these forms.
So to answer the question, yes, real numbers do get more exotic than numbers in the ring of periods or exponential periods like $e$ or $\pi$.
My understanding of statements of the form "$\Gamma$ takes values in the exponential periods at rational values" is that we would expect $\Gamma$ to take values at irrational arguments that are not exponential periods (although in general I expect proofs of such claims to be hard to come by).
So I would expect numbers like $\Gamma(\sqrt{2})$, $e^\pi$, $\zeta(\log 2)$ to be examples of computable numbers which are not exponential periods. But these would presumably not be considered “classical constants”.
Going further, even the computable numbers are countable, so most real numbers are not even computable, let alone in the ring of periods.
What you call the "degree of irrationality" of a number is the degree of its minimal polynomial over $\mathbb Q$, sometimes called the degree of the number over $\mathbb Q$. This concept is defined and explored in field theory.
Every real number is the root of a power series with integer coefficients.
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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.