I Need help creating an intuitive answer to the sum of $1(1!) + 2(2!) + 3(3!) +\cdots+ n(n!)$

Question

Given the sum $1+\sum_{i=1}^n i(i)! = (n+1)!$, is there an intuitive way to think about this sum? I understand the algebraic manipulation to get to that answer, and also how to use induction to prove the answer, but say could we use a combinatorial proof to get to the same answer? I cant seem to think of a way to think of anything that could intuitively explain this. I tried thinking about it as the number of ways to create license plates with restrictions on the number of different characters that are allowed to be used at any given spot in the license plate, whereas the right-hand side counts them all at once, but that didn't get me to an answer. Any help is greatly appreciated.

Answer 1

If we want to arrange $n+1$ distinct objects in a line ($(n+1)!$ ways in total), we can take the last object and either

• place it in any one of the first $n$ positions, after which we have $n!$ ways to permute the remaining objects ($n(n!)$ ways)
• place it in the last position

In the second case, we repeat the whole process with the first $n$ objects and free positions. If we place the second-last object in the second-last position, we continue with the first $n-1$ objects and free positions, etc. This recursion stops when we have only one object left to place in one position. The total number of arrangements that can be generated by this process is $$n(n!)+(n-1)(n-1)!+\cdots+1(1!)+1$$ Every permutation can be achieved this way, so the sum is equal to $(n+1)!$.