MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

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

0 like 0 dislike
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)

0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike