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We will start with the known identity that

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

and assume for contradiction that $e=\frac pq$. Then we have that $p=qe$, where $p$ is an integer. So then $q!e$ is also an integer number (specifically $p(q-1)!$). This means that

$$q!e=q!\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{q!}{n!}\in\mathbb{Z}.$$

Now, we can split this sum into two parts:

$$q!e=\sum_{n=0}^q \frac{q!}{n!}+\sum_{n=q+1}^\infty\frac{q!}{n!},$$

and note that each term in the first sum is an integer (because $n!|q!$ when $n\leq q$). On the other hand, in the second sum,

$$\sum_{n=q+1}^\infty\frac{q!}{n!}=\sum_{n=q+1}^\infty\frac{q!}{q!(q+1)\cdots(q+n)}\leq\sum_{n=q+1}^\infty\frac{1}{(q+1)^n}<1$$

where there are $n$ terms in $q!(q+1)\cdots(q+n)$. This gives us our contradiction. Since the first sum in $q!e$ are all integers and the second sum is positive and less than 1, $q!e\notin\mathbb{Z}$ and thus $e$ is irrational.

I can't specifically say anything that's wrong with it, but I still don't feel good about it. In particular, I'm skeptical about rewriting $n!=q!(q+1)\cdots(q+n)$ when taking a sum to infinity, as well as the inequality, but I can't specifically say what might be wrong about it.

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1 Answer 1

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You should have written $$\sum_{n=q+1}^\infty \frac{q!}{n!} = \sum_{n=1}^\infty \frac{q!}{q!(q+1)\cdots(q+n)} \le \sum_{n=1}^\infty \frac{1}{(q+1)^n} = \frac{1/(q+1)}{1 - 1/(q+1)} = \frac{1}{q} < 1$$ since $q > 1$. Note the index of summation shifts. The argument you made is sound.

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