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## 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.