Prove by induction that $\sum_{i=1}^n i!\times i=(n+1)!-1$ for all $n\in \mathbb{N}$

Question

So far I have,

If $P(n):\sum_{i=1}^n i!\times i=(n+1)!-1$, then

$P(1):\sum_{i=1}^1 i!\times i=1$ and $(1+1)!-1=1$ , so P(1) is true.

I know I now have to assume P(K) is true, such that $P(K):\sum_{i=1}^k i!\times i=(k+1)!-1$ and show that $P(k+1)$ takes the same form, so that $P(k)\implies P(k+1)$, but this is where I get stuck.

Answer 1

If $S(m)=\sum_{r=1}^mr!\cdot r$

$$S(n+1)=\sum_{r=1}^{n+1}r!\cdot r=S(n)+(n+1)\cdot (n+1)!$$ $$=(n+1)!-1+(n+1)\cdot (n+1)!$$

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

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Alternatively, w/o induction,

$\displaystyle r!\cdot r=r!\cdot(r+1-1)=(r+1)\cdot r!-r!=(r+1)!-r!$ which makes $S(m)$ a Telescoping Series

Answer 2

There's another way, it's called perturbation method developed by Donald Knuth. Consider $$ S_n = \sum_{k=1}^{n}k! $$ then (since $(k+1)! = k!(k+1)$ $$ S_{n+1} = S_{n} + (n+1)! = \sum_{k=1}^{n}(k+1)! = 1+ \sum_{k=1}^{n}k! + \sum_{k=1}^{n}k!k\\ =1 + S_n + \sum_{k=1}^{n}k!k $$ Obviously $S_n$ cancels out and you get your result.