Binomial sum with factorial
$n!(1+1+\frac1{2!}+\frac1{3!}+\cdots+\frac1{n!})=\lfloor n!e\rfloor$
Binomial sum with factorial
$n!(1+1+\frac1{2!}+\frac1{3!}+\cdots+\frac1{n!})=\lfloor n!e\rfloor$
$n!(1+1+\frac1{2!}+\frac1{3!}+\cdots+\frac1{n!})=\lfloor n!e\rfloor$