I have to prove that $e$ is irrational using this inequality

$$0<e-\sum_{k=0}^n\frac1{k!}<\frac1{n\cdot n!}$$

The exercise leave the hint "prove by contradiction". I know too that $2<e\le 3$.

What I tried is set $e=p/q$ for $p,q\in\Bbb N$, and $\sum 1/k!=A/n!$ where $A\in\Bbb N$. Then I written


but I dont get any idea from here. Indeed I dont know exactly what to do here, I never used a inequality to prove the irrationality of a number. Can you give me some hint (or solution)? Thank you.

P.S.: I dont know exactly what kind of tags I have to use for this question.

  • $\begingroup$ do you know the series representation of $e$? $\endgroup$ – tired Oct 19 '16 at 10:56
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    $\begingroup$ Be careful here: $A$ still depends on $n$, so it would be better to write $A(n)$. Otherwise, if $A$ is a fixed integer, the contradiction would be obvious. :) $\endgroup$ – Dirk Oct 19 '16 at 10:56
  • $\begingroup$ @tired, yes, I know that $e=\sum\frac1{k!}=\lim (1+\frac1n)^n$ $\endgroup$ – Masacroso Oct 19 '16 at 11:00
  • $\begingroup$ $$e-\sum_{0}^n \frac{1}{k!}=\sum_{n+1}^{\infty} \frac{1}{k!}=\frac{1}{(n+1)!}\left[1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+...\right]<\frac{1}{(n+1)!}\left[1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+...\right]=\frac{1}{(n+1)!}\frac{n+1}{n}=\frac{1}{n n!}$$ $\endgroup$ – tired Oct 19 '16 at 11:21

Hint: If $e=\frac{p}{q}$, take $n=q$ in the given inequality and clear denominators.

  • $\begingroup$ Sorry but Im unable to solve from this hint. I get $$0<q!\left(p-q\sum\frac1{k!}\right)<1$$ and I know that $p\in(2q,3q)$ and $\sum\frac1{k!}\in(2,3)$. $\endgroup$ – Masacroso Oct 19 '16 at 11:58
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    $\begingroup$ @Masacroso, you're almost there. Just note that the number between $0$ and $1$ is an integer! $\endgroup$ – lhf Oct 19 '16 at 11:59

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