Yesterday evening, I wrote this proof that e is irrational. I showed it to my maths teacher at school and he thought it was correct. I wondered if anyone else could tell me if it was valid and if anyone else has proven it in this way before.
If e is rational, then it must be possible to write it as $\frac ab$ where a and b are positive integers. Therefore it must be possible to write $e=\frac {a\times b!}{b\times b!}$ and since $a\times b!$ is an integer, there is must be some integer, c, such that $e=\frac {c}{b\times b!}$.
$$ e = \frac {1}{0!} + \frac {1}{1!} +...+ \frac {1}{b!} + \frac {1}{(b+1)!} + \frac {1}{(b+2)!} + \frac {1}{(b+3)!} + ...$$
$(b+1)\times(b+1)! < (b+2)!$ and so $\frac {1}{(b+1)\times(b+1)!} > \frac {1}{(b+2)!}$. Similarly, $\frac {1}{(b+1)^2\times(b+1)!} > \frac {1}{(b+3)!}$. Therefore,
$$e<\frac {1}{0!} + \frac {1}{1!} +... +\frac {1}{b!} + \frac {1}{(b+1)!} + \frac {1}{(b+1)\times(b+1)!} + \frac {1}{(b+1)^2\times(b+1)!} + ...$$
The part of this after $\frac {1}{b!}$ forms a geometric series and since the ratio of one term to the next has a magnitude of less than 1, this series is convergent. Since the infinite sum of a convergent geometric series is $\frac {a}{1-r}$ where a is the first term and r is the ratio of one term to the next, the sum of this geometric series is $\frac {1}{b\times b!}$. Therefore, $$e<\frac {1}{0!} + \frac {1}{1!} +... +\frac {1}{b!} + \frac {1}{b\times b!}$$ Since each of the terms in the first infinite series is positive,$$e>\frac {1}{0!} + \frac {1}{1!} +... +\frac {1}{b!}=\frac{I}{b\times b!}$$ for some integer I. Therefore, $$\frac{I}{b\times b!}<e<\frac{I}{b\times b!}+\frac {1}{b\times b!}$$ $$\frac{I}{b\times b!}<e<\frac{I+1}{b\times b!}$$ However, in order for e to be rational, we said that there must be some integer, c, such that $e=\frac {c}{b\times b!}$. But there is no integer between $I$ and $I+1$. Therefore, by contradiction, e must be irrational.