Close form of $1+\sum_{j=1}^n j! j$

Moderator: elim

Close form of $1+\sum_{j=1}^n j! j$

Find the close form of $\displaystyle{1+\sum_{j=1}^n j! j }$
elim

Posts: 64
Joined: Mon Feb 08, 2010 8:03 pm

Re: Close form of $1+\sum_{j=1}^n j! j$

$\displaystyle{1+\sum_{j=1}^n j! j = 1+\sum_{j=1}^n \left((j+1)! -j! \right ) = \sum_{j=1}^{n+1} j! - \sum_{j=1}^n j! = (n+1)! }$
elim

Posts: 64
Joined: Mon Feb 08, 2010 8:03 pm