Proof that e is irrational using Taylor series

[Incand]

Homework Statement

Using the equality $e = \sum_{k=0}^n \frac{1}{k!} + e^\theta \frac{1}{(n+1)!}$ with $0< \theta < 1$, show the inequality $0 < n!e-a_n<\frac{e}{n+1}$ where $a_n$ is a natural number.
Use this to show that $e$ is irrational.
(Hint: set $e=p/q$ and $n=q$)

Homework Equations

N/A

The Attempt at a Solution

I think I manage to do the first part. Multiply both sides with $n!$
$n!e = \sum_{k=0}^n \frac{n!}{k!} + \frac{e^\theta}{n+1}$
Setting $\theta = 0$ and $\theta=1$ respectively we have the inequality
$0 < n!e-a_n<\frac{e}{n+1}$ with $a_n = \frac{n!}{k!}$.

For the second part assume $e= \frac{p}{q}$ where $p,q$ are natural numbers without a common divisor (except 1).
Set $n=q$ in the above inequality
$0 < q!\frac{p}{q}-a_q < \frac{p}{q(q+1)}$
Since $q,p$ and $q+1$ is positive we can multiply (divide) by these
$0 < (q+1)(q!-qa_q/p)< 1$
My hopes at this point would be to have an integer in the middle and hence get a contradiction (since there is no integers between 0 and 1), but that $p$ is in the way.

 

[Samy_A]

Homework Statement

Using the equality $e = \sum_{k=0}^n \frac{1}{k!} + e^\theta \frac{1}{(n+1)!}$ with $0< \theta < 1$, show the inequality $0 < n!e-a_n<\frac{e}{n+1}$ where $a_n$ is a natural number.
Use this to show that $e$ is irrational.
(Hint: set $e=p/q$ and $n=q$)

Homework Equations

N/A

The Attempt at a Solution

I think I manage to do the first part. Multiply both sides with $n!$
$n!e = \sum_{k=0}^n \frac{n!}{k!} + \frac{e^\theta}{n+1}$
Setting $\theta = 0$ and $\theta=1$ respectively we have the inequality
$0 < n!e-a_n<\frac{e}{n+1}$ with $a_n = \frac{n!}{k!}$.

For the second part assume $e= \frac{p}{q}$ where $p,q$ are natural numbers without a common divisor (except 1).
Set $n=q$ in the above inequality
$0 < q!\frac{p}{q}-a_q < \frac{p}{q(q+1)}$
Since $q,p$ and $q+1$ is positive we can multiply (divide) by these
$0 < (q+1)(q!-qa_q/p)< 1$
My hopes at this point would be to have an integer in the middle and hence get a contradiction (since there is no integers between 0 and 1), but that $p$ is in the way.

From $0 < q!\frac{p}{q}-a_q < \frac{p}{q(q+1)}$, you get
$0<p(q-1)!-a_q<\frac{e}{q+1}$.
I you could show that $\frac{e}{q+1}<1$, you are done.

 

[Incand]

If you could show that $\frac{e}{q+1}<1$, you are done.

Thanks! We know that $2 < e < 3$ (this is easy to prove using the limit definition of e). If $q \ge 2$ then the inequality is obviously true since $e < 3$ and hence we have a contradiction.

The case for $q=1$ means we have a natural number so it's enough to show it for $q \ge 2$ as above.