Exercise 6.3.8 ($e$ is irrational)

Question

Recalling that the mathematical constant $e$ has value $\sum_{j=0}^\infty 1/j!$, prove that
$$\lfloor (k!)e \rfloor = k! \sum_{j=0}^k \frac{1}{j!} < (k!) e.$$
Hence prove that $e$ is irrational.

Richard Ganaye · Accepted Answer

\begin{proof} 
We note that for all positive integer $k$,
\begin{align*}
\sum_{j = k+1}^\infty \frac{1}{j!} &= \frac{1}{(k+1)!} \left( 1 + \frac{1}{k+2} + \frac{1}{(k+2)(k+3)}+ \cdots + \frac{1}{(k+2)(k+3)\cdots j} + \cdots ight)\
&< \frac{1}{(k+1)!} \left( 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{j-k-1}} + \cdots ight) = \frac{2}{(k+1)!}.\
\end{align*}
Therefore
$$ \sum_{j = k+1}^\infty \frac{k!}{j!} < \frac{2}{k+1} \leq 1.$$
Then 
$$N = \sum_{j = 0}^k\frac{k!}{j!} < k! e = \sum_{k=0}^\infty \frac{k!}{j!} < \sum_{j = 0}^k\frac{k!}{j!}  + 1 = N + 1.$$
Since $N$ is an integer, this shows that
$$\lfloor k! e floor = \sum_{j = 0}^k\frac{k!}{j!} < k! e.$$
So for every positive integer $k$, $\lfloor k! e floor  
e k! e$. By Problem 7, $e$ is irrational.
\end{proof}

Exercise 6.3.8 ($e$ is irrational)

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Exercise 6.3.8 ($e$ is irrational)

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