How does this series yield an irrational function

Question

It is well known that the euler number $e$ is irrational. It is also well known that the Taylor expansion of $e$ can be represented as $$e=\sum_{k=0}^{\infty}\frac{1}{k!}$$ Now, when we look at the term $$T(k)=\frac{1}{k!}\quad\forall k\in\mathbb{Z}$$ We realize that $T(k)$ has to be a rational number for all $k$. How can a sum of rational numbers yield an irrational number $e$? I am not a mathematician, hence all and any help is appreciated.

Answer 1

Every irrational number is an infinite sum of rational numbers

For example $$\sqrt 2 =1+.4+.01+.004+.....$$ The magic word is infinite sum.

Answer 2

Take the square root of $2$, which is known to be irrational; it's roughly $1.41428\ldots$. I can write that as $$ 1 + \frac{4}{10} + \frac{1}{100} + \frac{4}{1000} + \frac{2}{10000} + \frac{8}{100000} + \ldots $$ When I do so, you can see that each individual term is a rational number.

Now you might be thinking "but I know that when I add two rationals, I get a rational:" $$ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} $$ ... so why isn't the larger sum still rational?

The answer is that any finite part of it is rational. The first three terms, for instance, give the rational $$ \frac{141}{100}. $$

But that doesn't mean that an infinite sum must be rational as well, and indeed, that's not true.

Let me work by analogy with another notion: any finite sum of numbers is finite. But an infinite sum of numbers is not necessarily finite, as $$ 1 + 1 + 1 + \ldots $$ shows.

So just because you've proven some property works for pairs of things or finite collections, you don't necessarily know that it works for infinite things.

That subtle fact is a large part of what the third portion of most serious calculus courses is all about -- the "sequences and series" part. So to really understand it, you've got some work to do, alas.

Answer 3

If the sum was finite, then of course we could add all the rational numbers into a fraction with the result being rational. But since the sum has infinitely many terms, there is a possibility that the sum does not approach something we can represent as the ratio of two integers. In that case, it is irrational.

If I remember correctly, one of the classic proofs of the irrationality of e does not need more than a bit of calculus to follow.

Answer 4

That's quite a normal situation. When you “sum a series”, you don't actually do a sum, but take a limit of finite sums: $$ \sum_{n=0}^{\infty} a_n=\lim_{n\to\infty}(a_0+a_1+\dots+a_n) $$ Even if each $a_n$ is rational, the limit is not a (finite) sum of rational numbers, so there's no compelling reason for it to be rational.

The Taylor-Lagrange theorem applied to the exponential function says that, for any integer $k$, there exists $c_k\in(0,x)$ such that $$ e=\sum_{n=0}^k\frac{1^n}{n!}+\frac{e^{c_k}}{(k+1)!} $$ because the $n$-th derivative of $e^x$ is $e^x$.

Suppose, to the contrary, that $e=a/b$, for some positive integers $a$ and $b$. Take $k>\max\{b,3\}$; then $$ e=\sum_{n=0}^k\frac{1^n}{n!}+\frac{e^{c_k}}{(k+1)!}< \sum_{n=0}^k\frac{1^n}{n!}+\frac{3}{(k+1)!} $$ because $c_k<1$ and $e<3$. Thus $$ 0<\left(e-\left(1+1+\frac{1}{2!}+\dots+\frac{1}{k!}\right)k!\right)< \frac{3}{(k+1)!}k!=\frac{3}{k+1} $$ However, the term $$ e-\left(1+1+\frac{1}{2!}+\dots+\frac{1}{k!}\right)k! $$ is integer by assumption, as $k!$ is a multiple of $b$. On the other hand $$ \frac{3}{k+1}<\frac{3}{3+1}=\frac{3}{4} $$ which is a contradiction.

Thus $e$ is irrational and it's the very structure of $$ e=\sum_{n=0}^\infty\frac{1}{n!} $$ that makes it so.