Let $\displaystyle x=1/1!+1/1!2!+1/1!2!3!+...$ Prove that x is an irrational number.

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- Nov 23rd 2008, 07:44 PMmathman7irrational number?
Let $\displaystyle x=1/1!+1/1!2!+1/1!2!3!+...$ Prove that x is an irrational number.

- Nov 24th 2008, 12:15 AMKiwi_Dave
Hopefully someone will give you a more elegant soln than mine!

Assume that x is rational so the integers N and M exist such that:

x=N/M

Therefore, x*(1!2!3!.....M!) must be an integer.

But

$\displaystyle x*(1!2!3!.....M!)=(1/1!+1/1!2!+1/1!2!3!+...)*(1!2!3!.....M!)$

Now the first M terms of this are integers. Define S to be the integer that is the sum of the first M integers then:

$\displaystyle x*(1!2!3!.....M!)=S+(1/(M+1)!+1/(M+1)!(M+2)!+...)$

Now provided M > 1

$\displaystyle 0<(1/(M+1)!+1/(M+1)!(M+2)!+...)<(1/2^2+1/3^2+...)=\pi^2/6-1<1$

Since this term is between 0 and 1 it is not an integer and we have a contradiction with our statement that N is an integer and hence x must be irrational.