Proving $\sum_{k=1}^{n} k(k!) = (n+1)! - 1$ by induction

Question

How do you prove this using induction? $$\sum_{k=1}^{n} k(k!) = (n+1)! - 1$$ for all positive integers n

Base Case: $1(1!) = 1$ and $(2!) - 1 = 1$, so this works for$ n = 1$.

How do I show the $n + 1$ case?

Answer 1

$$S=\sum_{k=0}^{n} k ~ k!= \sum_{k=0}^{n} [(k+1) -1]~ k!= \sum_{k=0}^{n}[(k+1)!- k!]=(n+1)!-1.$$ In the last step we have done telescoping summation.

Answer 2

Actually using induction.

If $\sum_{k=1}^{n} k(k!) = (n+1)! - 1 $ then

$\begin{array}\\ \sum_{k=1}^{n+1} k(k!) &=\sum_{k=1}^{n} k(k!)+(n+1)(n+1)!\\ &=(n+1)! - 1+(n+1)(n+1)!\\ &=(n+2)(n+1)!-1\\ &=(n+2)!-1\\ \end{array}\\ $

which is what you want to prove.

Answer 3

If $\sum_{k=1}^{n} k(k!) = (n+1)! - 1$

Then $(n+1)(n+1)! + \sum_{k=1}^{n} k(k!) = (n+1)(n+1)! + (n+1)! -1 $

$\sum_{k=1}^{n+1} k(k!)= (n+1)![(n+1) + 1] - 1= (n+1)!(n+2)-1 =(n+2)! - 1$