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

Question

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

This proposition seem to be true

First step $P(1)$

$1=2!-1$

Second step assume $P(k)$

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

Third step $P(k+1)$

The area of difficulty for me.

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

$\sum_{i=1}^{k+1}(i)(i!)=(k+1)(k+1)!+\sum_{i=1}^{k}(i)(i!)$

which using the asumption is

$\sum_{i=1}^{k+1}(i)(i!)=(k+1)(k+1)!+\sum_{i=1}^{k}(i)(i!)$

which using the induction hypothesis is

$(k+1)(k+1)!+(k+1)!-1$

but I am unsure what to do next.

Answer 1

Since you have already done the base case assume it is true for $n$ we will show it for $n+1$:

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

Answer 2

Another way, without induction. It's called a perturbation method in Concrete Mathematics. Define $S_n = \sum_{k=1}^{n}k!$. $$ S_n + a_{n+1} = \sum_{k=1}^{n}k! + (n+1)! = \sum_{k=1}^{n}(k+1)!+1=\sum_{k=1}^{n}k!(k+1) + 1 = \sum_{k=1}^{n}k!k +\sum_{k=1}^{n}k! +1 $$ Clearly $\sum_{k=1}^{n}k!$ terms on both sides cancel out so you get your result: $$ \sum_{k=1}^{n}k!k = (n+1)!-1 $$