sum of n factorials?

• 1!+2!+---+n!
  = n!(1/n!+ 2/n! +......+1/n + 1)
  so if the value of n is large then all the terms in the bracket except " 1/n and 1 " can be assumed to be zero.
  
  = n! (n+1)/n = (n+1)! / n is the answer.
  
  1!+2!+---+n! = (n+1)!/n
  
• 10 years agoHelpfull: Yes(2) No(0)
• 
  0!+1!+2!+---+n!;
• 10 years agoHelpfull: Yes(0) No(1)