1

Maybe this question is too dumb to be asked, but it's really bugging me so I decide to ask it anyway. I hope you bear with me.

Okay, it's known that both sides of the following series equal.

$$\pi=4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\tag1$$

and

$$e=\sum_{n=0}^{\infty}\frac{1}{n!}\tag2$$

We all agree at this point. Now, each terms in $(1)$ and $(2)$ is a rational number. We all agree without a doubt. The sum of rational numbers is always a rational number. We agree again. Hence it follows that $\pi$ and $e$ must be rational numbers. However, it contradicts the well-known facts that both $\pi$ and $e$ are irrational numbers. So, where is my mistake?

L Lawliet
  • 809

2 Answers2

13

An "infinite sum" is not a sum.

An infinite sum is the limit of a sequence :

$$\sum_{n=0}^\infty=\lim_{N\to\infty} \sum_{n=0}^N.$$

An the limit of a sequence of rational numbers is not necessarily rational.

E. Joseph
  • 14,843
6

Take any known irrational number, $x$. You can always represent $x$ by the sum of an infinite number of rational numbers.

For example, take the decimal representation of $x$:

$x = 5.1938527\ldots$

and then each term in the sum could form one of the decimal digits:

$x =5 + \frac{1}{10} + \frac{9}{100} + \frac{3}{1000} + \cdots$

Therefore the sum of an infinite number of rationals is not always rational.