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...
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1$\begingroup$ $$k!e = k!\sum_{j=0}^k \frac{1}{j!} + \sum_{j=k+1}^\infty \frac{k!}{j!}.$$ See that the second sum is smaller than $1$ and the first sum an integer. $\endgroup$– Daniel FischerMay 6, 2014 at 13:38
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1$\begingroup$ Ohhhh!! I should sleep... I can't believe I didn't see this. $\endgroup$– phoenixMay 6, 2014 at 13:38
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1 Answer
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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: $$[k!e]=k!+\frac{k!}{2}+\cdots+1$$ $$[k!e]=k!\sum_{j=0}^k \frac{1}{j!}$$