Proving $\sum_{k=1}^n kk!=(n+1)!−1$

Question

Ok, so I'm trying to prove statement in the header. I have read the following discussion on it, but I can't seem to follow it all the way through:

I like mfl's answer, but I get hung up on the last step. They say:

and we need to show

$$\sum_{k=1}^{n+1} kk!=(n+2)!−1.$$

Just write

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

How do they get from the first step stated above, to the following step? I'm stuck.

If we would have extended the summation to (n+1) terms what would be the (n+1)th term? — Rohan Shinde, Jan 18 '18 at 03:43
No it isn't. It would be (n+1)(n+1)!. You must see the general term of the summation and I just asked you the (n+1)$\mathbf {th}$ term not the sum of (n+1) terms. I hope you understood why mfl had written the (n+1)(n+1)! In addition to the summation ( because it is simply the (n+1)th term i.e.last term of the summation in the LHS of last step you mentioned above.). He just wrote that term out of the summation to apply the induction. — Rohan Shinde, Jan 18 '18 at 05:18

score 1 · Answer 1 · answered Jan 18 '18 at 03:41

1

$$\sum_{k=1}^n kk!= \sum_{k=1}^{n-1} kk!+nn!$$

And by induction we know $\sum_{k=1}^{n-1} kk!= n!-1$

Combining we get: $$\sum_{k=1}^{n-1} kk!+nn! =n!-1+nn! = n!(1+n)-1=(n+1)!-1$$

answered Jan 18 '18 at 03:41

score 1 · Answer 2 · answered Jan 18 '18 at 04:14

1

Another way is to write : $$\sum_{k=1}^n k k!=\sum_{k=1}^n (k+1 -1) k!=\sum_{k=1}^n (k+1)!-k!=(n+1)!-1$$

answered Jan 18 '18 at 04:14

Jaideep Khare

2 Answers2