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Show that the base of the natural logarithm, $e$ as defined below, is irrational. You may assume the sequence converges.
$$
e=\lim _{n \rightarrow \infty}\left(\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{n !}\right) .
$$
in Mathematics by Bronze Status (9,632 points) | 37 views

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Proof. Let
$$
e_{n}=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{n !} .
$$
We then have, if $n$ is a natural number,
$$
\begin{aligned}
0<e-e_{n} &=\lim _{N \rightarrow \infty}\left(\frac{1}{(n+1) !}+\frac{1}{(n+2) !}+\frac{1}{(n+3) !}+\cdots+\frac{1}{(n+N) !}\right) \\
&=\frac{1}{(n+1) !} \lim _{N \rightarrow \infty}\left(1+\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+\cdots+\frac{1}{(n+2)(n+3) \cdots(n+N)}\right) \\
& \leq \frac{1}{(n+1) !} \lim _{N \rightarrow \infty}\left(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^{2}}+\cdots+\frac{1}{(n+2)^{N-1}}\right) \\
&=\frac{1}{(n+1) !} \lim _{N \rightarrow \infty} \frac{1-\frac{1}{(n+1)^{N}}}{1-\frac{1}{n+1}}=\frac{1}{(n+1) !} \frac{n+1}{n}=\frac{1}{n \times n !}
\end{aligned}
$$
Collecting this information, we get
$$
0<e-e_{n}<\frac{1}{n \times n !}
$$

Now if $e$ is a rational number, then $e=p / q$ with natural numbers $p$ and $q$. We have, from $(1)$,
(2)
$$
0<e-e_{q}<\frac{1}{q \times q !} \text { which gives } 0<q !\left(e-e_{q}\right)<\frac{1}{q} \text { . }
$$
We note here that $e \times q !$ is certainly a natural number and so is
$$
q ! \times e_{q}=q ! \times\left(\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{q !}\right) .
$$
Hence $q !\left(e-e_{q}\right)$ is the difference between two natural numbers and hence an integer. But this quantity lies strictly between zero and one as in shown in (2) which is certainly an absurdity. Hence $e$ must be irrational.
by Bronze Status (9,632 points)

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