I found just one question similar to this, but it had been edited, so hopefully this isn't asked too often.
Given the formulas via infinite sums for expressing $e$ and $\pi$... $$ e = \sum_{n=0}^\infty \frac{1}{n!} = 1+ 1 + \frac{1}{2} + \frac{1}{6} + ...$$ $$ \pi = 4\sum_{n=0}^\infty \frac{(-1)^{n}}{2n+1} = 4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7} + ... $$
Since every number in each series is rational, and the sum of two rational numbers is also rational, doesn't that imply that $e$ + $\pi$ is rational?
I realize this may seem like a simple argument, but the only way I can imagine it being wrong is if there is a principle I am missing regarding adding two infinite sums.
An infinite sum of rationals may be either rational or irrational. In fact, any real number, whether rational or not, is an infinite sum of rationals $$n+\frac{d_1}{10}+\frac{d_2}{100}+\frac{d_3}{1000}+\cdots\ :$$ this is exactly what its decimal representation means.
The problem is that an "infinite sum" is not, strictly speaking, a sum. It is a sum together with a limit: by definition, $$\sum_{n=1}^\infty a_n$$ means $$\lim_{N\to\infty}\sum_{n=1}^N a_n\ .$$ The fact that a limit is involved changes a number of the usual rules regarding addition.