I was looking at some books that have solutions to this, but I'm wondering which of this is correct, like kk will be the same as k(k)
One is:
Claim: For all $ n \geq 1, \sum_{k=1}^{n} k k !=(n+1) !-1 $.
Proof. Let $S(n) $ denote the statement $[ S(n): \sum_{k=1}^{n} k k !=(n+1) !-1 . ]$
Base step $ (n=1): S(1) $ is true because $ 1=2 !-1 $.
Inductive step: For some fixed $ \ell \geq 1 $, assume the inductive hypothesis $ S(\ell) $ to be true where $[ S(\ell): \sum_{k=1}^{\ell} k k !=(\ell+1) !-1 . ]$
To be shown is that $ S(\ell+1) $ follows, where $[ S(\ell+1): \sum_{k=1}^{\ell+1} k k !=(\ell+2) !-1 . ]$
Starting with the left-hand side of $ S(\ell+1) $, $[ \begin{aligned} \sum_{k=1}^{\ell+1} k k ! & =\sum_{k=1}^{\ell} k k !+(\ell+1)(\ell+1) ! \\ & =[(\ell+1) !-1]+(\ell+1)(\ell+1) ! \\ & =(\ell+1) !(1+\ell+1)-1 \\ & =(\ell+2) \cdot(\ell+1) !-1 \\ & =(\ell+2) !-1, \end{aligned} ]$ (by definition of $ \Sigma $ ) (by $ S(\ell) $ ) we see that the right-hand side of $ S(\ell+1) $ follows. This completes the inductive step.
Thus, by mathematical induction, the statement $ S(n) $ is true for all $ n \geq 1 $.
which I don't quite understand but came across this:
If $ \sum_{k=1}^{n} k(k !)=(n+1) !-1 $ then $ \begin{aligned} \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{aligned} $
So with all I think it will be something more like this? Or am I wrong??? Base case: For $ n = 1 $, the left-hand side is $ 1*1! = 1 $ and the right-hand side is $ (1+1)! - 1 = 2! - 1 = 1 $. So the equation holds for $ n = 1 $.
Inductive step: Assume the equation holds for some $ n $, i.e., $ \sum_{k=1}^{n} k(k!) = (n+1)! - 1 )$. We need to show that the equation also holds for $( n+1 ), i.e., ( \sum_{k=1}^{n+1} k(k!) = (n+2)! - 1 $.
Starting with the left-hand side of the equation for \$ n+1 $:
[ \begin{aligned} \sum_{k=1}^{n+1} k(k!) & = \sum_{k=1}^{n} k(k!) + (n+1)((n+1)!) \\ & = (n+1)! - 1 + (n+1)((n+1)!) \quad \text{(by the inductive hypothesis)} \\ & = (n+2)(n+1)! - 1 \\ & = (n+2)! - 1 \end{aligned} ]
Which is the right-hand side of the equation for ( n+1 ). So the equation holds for ( n+1 ) if it holds for ( n ), completing the induction.