how to prove that $\sum_{k=1}^{m}k!k=(m+1)!-1$ without induction?

Question

how to prove that $$\sum_{k=1}^{m}k!k=(m+1)!-1$$ without induction ?

my only try is to put $k!=\Gamma(k+1)$ then use geometric series with some steps but I got complicated integral

If any one can solve it using my way or similar way using calculus technique

First of all, all you can ever do with this sort of problem is hide the induction. You can't even define the sum "without induction." — Thomas Andrews, Mar 12 '14 at 23:12

score 12 · Accepted Answer · answered Mar 12 '14 at 23:13

12

$$ \sum_{k=1}^{m}k!k \\ \sum_{k=1}^{m}k!(k+1-1) \\ \sum_{k=1}^{m}k!(k+1)-k! \\ \sum_{k=1}^{m}(k+1)!-k! \\ = 2!-1!+3!-2!+4!-3!+\cdots + (m+1)!-m! \\ =(m+1)!-1 $$

answered Mar 12 '14 at 23:13

Priyatham

2,607

oh my god but I need 4 minutes more – Mar 12 '14 at 23:17
Telescopy is induction in disguise. But as Thomas Andrews pointed, it is impossible to prove this without induction. This argument is kind of nice, though. +1. – chubakueno Mar 12 '14 at 23:31
@chubakueno I agree and thanks. – Priyatham Mar 12 '14 at 23:32

score 6 · Answer 2 · answered Mar 12 '14 at 23:20

There is a combinatorial approach.

The number $(m+1)!-1$ counts the number of non-identity permutations of $\{1,2,\dots,m+1\}$.

On the other hand $k\cdot k!$ counts the number of such permutations that fix all elements greater than $k+1$ but not $k+1$.

how to prove that $\sum_{k=1}^{m}k!k=(m+1)!-1$ without induction?

2 Answers2