## Sum of Series

Sum of Series
Find the sum of the series $1(1!)\!+2(2!)\!+3(3!)\!+\cdots+n(n!)$
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
$1(1!)\!+2(2!)\!+3(3!)\!+\cdots+n(n!)\!=[2(1!)\!-1!]+[3(2!)\!-2!]+[4(3!)\!-3!]+\cdots+[n(n-1)!\!-(n-1)!]+[(n+1)(n!)\!-n!]$
$=2!\!-1!\!+3!\!-2!\!+4!\!-3!\!+\cdots+n!\!-(n-1)!\!+(n+1)!\!-n!$
Note that $2(1!)\!=2!,3(2!)\!=3!,4(3!)\!=4!$, and so on
$=2!\!+3!\!+4!\!+\cdots+n!\!+(n+1)!\!-1!\!-2!\!-3!\!-4!\!-\cdots-(n-1)!\!-n!$
$=(n+1)!\!-1!$
$=(n+1)!\!-1$

Answer: $(n+1)!\!-1$