Irrationality of e and farey fractions

Question

How do we go about proving that $$[k! e] = k! \sum_{j=0 -> k} \frac{1}{j!}$$ I know that we could write $$e = \sum_{j=0 -> \infty} \frac{1}{j!}$$ But I don't see how that's going to help in the proof...

Answer 1

Sketch proof: $$k!e=k!+\frac{k!}{2}+\cdots+1+k!\sum_{j=k+1}^\infty\frac{1}{j!}$$ If with $[\ ]$ you mean the integer for proof is: $$k!\sum_{j=k+1}^\infty\frac{1}{j!}<1$$ So: $$[k!e]=k!+\frac{k!}{2}+\cdots+1$$ $$[k!e]=k!\sum_{j=0}^k \frac{1}{j!}$$