proving "e" irrational using convergent series

307 Views Asked by At

So my math professor in college gave us this problem:

$$U_n=\sum_{k=1}^{k=n}\frac{1}{k!}$$

$$V_n=U_n+ \frac{1}{n\cdot n!}$$

The problem has two questions:

1- Prove that the two series are adjacent (easy)

2- Let $e$ be their limit, prove that $e$ is irrational.

I have no idea how to start the second part, since I don't have a professor that does the problems the first professor gives us, so I have a lack in reasoning skills, I tried proof by contradiction, but no result.

also I'm new to the site so I don't know how to use the symbols so sorry for that.

1

There are 1 best solutions below

0
On BEST ANSWER

The letter $e$ is used for $\sum_{k=0}^{\infty}1/k!$ where, by definition, $0!=1. $ So $e=1+\lim_{n\to \infty} U_n$. If $e$ were rational then $n!e$ is an integer for some $n>0.$ (For example if $e=a/b$ with $a,b\in Z^+$ let $n=b$.)$$\text { But...... } n!e=A_n+B_n ,$$ $$\text {where } A_n=n!\sum_{k=0}^{k=n}1/k!=\sum_{k=0}^{k=n}(n!/k!)\in Z$$ and $$0<B_n=n!\sum_{k=n+1}^{\infty}1/k!=\sum_{k=n+1}^{\infty}n!/k!<\sum_{j=0}^{\infty}(n+1)^{-1}(n+2)^{-j}=(n+2)(n+1)^{-2}<1$$ so $B_n$ is not an integer.