Finding value of the following $\displaystyle \bigg \lfloor \frac{2017!}{2016!+2015!+\cdots \cdots + 2!+1!}\bigg \rfloor $
Attempt : $$2017! = 2017\cdot 2016! = (1+1+1+\cdots \cdots +1)2016!>2016!+2015!+2014!+\cdots +2!+1!$$
Could some help me how to solve it, thanks
Dividing by $2016!$ you get $$\frac{2017}{1+\frac{1}{2016}+\frac{1}{2016\cdot 2015}+\cdots+\frac{1}{2016!}}$$ Now consider $$\frac{2017}{1+\frac{1}{2016}}=2016$$ This implies that the full expression is smaller then $<2016$,and we have that $$\frac{2017}{1+\frac{1.1}{2016}}>2015$$ The sum of the rest of the terms is $$\frac{1}{2016\cdot 2015}+\cdots+\frac{1}{2016!}<\frac{1.1}{2016}$$ This could be seen easily since each term gets around $10^{-3}$ times smaller. Hence the result is $2015$
NOTE: The $1.1$ was arbitrary,it's easy to prove and easily satisfies the inequality
