Proving that $\pi$ and $e$ are rational numbers [duplicate]

This question already has an answer here:

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?

marked as duplicate by Dietrich Burde, Watson, Michael Hoppe, Dan Rust, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 17 '16 at 10:29

• "The finite sum of rational numbers is always a rational number" – Kevin Quirin Jun 17 '16 at 9:08
• This, as commented above, is one of the most important and sharp difference between finite and infinite in these matters: infinite (converging, of course) sums of rationals don't necessarily are rationals – DonAntonio Jun 17 '16 at 9:14
• See also this question. – Dietrich Burde Jun 17 '16 at 9:16
• Perhaps $\sqrt{2} = \dfrac{1}{1}+\dfrac{4}{10}+\dfrac{1}{100}+\dfrac{2}{1000}+\cdots$ as simpler counterexample – Henry Jun 17 '16 at 9:17

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.

• This is as simple, short and accurate as possibly expected (by me, at least). Very nice. +1 – DonAntonio Jun 17 '16 at 9:21

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.