Summation involving a factorial: $1 + \sum_{j=1}^{n} j!j$ [duplicate]

Solution 1:

This is a simple induction. Use $(n+1)! = (n+1) \cdot n! = n\cdot n! + n!$.

Added: I don't see any better way than to play around with the formula. Rearrange this and you have $j \cdot j! = (j+1)! - j!$. But then you have \begin{align*} \sum_{j=1}^{n} j \cdot j! & = \sum_{j=1}^{n} [ (j+1)! - j!] \\ & = [2! - 1!] + [3! - 2!] + \cdots + [n! - (n-1)!] + [(n+1)! - n!] \end{align*} and you see that everything cancels, except the terms $- 1!$ from the first summand and $(n+1)!$ from the last summand, hence the sum must be equal to $(n+1)! - 1$.

Edit 2

Here's the argument: We want to prove that the following statement $T(n)$ holds for all $n \in \mathbb{N}$: \[ (n+1)! - 1 = \sum_{j=1}^{n} j \cdot j!. \] For $n = 1$ we have the statement $T(1)$: \[ 1 = (1+1)! - 1 = \sum_{j=1}^{1} j \cdot j! = 1\cdot 1! = 1, \] so this is ok. Assume that $T(n)$ holds. We want to prove $T(n+1)$: \[ (n+2)! - 1 = \sum_{j=1}^{n+1} j \cdot j!. \] Start with the right hand side: \[ \sum_{j=1}^{n+1} j \cdot j! = (n+1)\cdot (n+1)! + \sum_{j=1}^{n} j \cdot j! \] But the last sum is equal to $(n+1)! - 1$ by our assumption that $T(n)$ is true, so \begin{align*} \sum_{j=1}^{n+1} j \cdot j! & = (n+1)\cdot (n+1)! + (n+1)! - 1 \\ & = [(n+1) + 1]\cdot(n+1)! - 1 = (n+2) \cdot (n+1)! - 1\\ & = (n+2)! - 1, \end{align*} so $T(n+1)$ holds as well.

Solution 2:

There is a nice interpretation of this identity in terms of uniqueness of representation in factorial base. But to answer your comment to Theo Buehler's answer, telescoping sequences are just a thing that you should be aware of and try to look for, and the identity Theo used is actually equivalent to what Wolfram told you, so...