How to simplify the summation kk! without using induction?

Question

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

I know the answer is (n+1)!-1..I can solve this question using principle of mathematical induction...but I would like to know if there is any other alternative approach

Answer 1

Set a permutation $\pi = \pi_1 \ldots \pi_{n+1}$ in $S_{n+1}$.

Now let $m = m(\pi)$ be the maximal index such that $\pi_1 = 1, \pi_2 = 2, \ldots, \pi_m = m$.

The number of permutations such that $m(\pi) = m$ for $m < n$ is $(n-m) (n-m)!$: now $n-m$ is the number of choices for $\pi_{m+1} \neq m+1$, and $(n-m)!$ is the number of permutations of the remaining $n-m$ numbers.

No permutation can satisfy $m(\pi) = n$, and there is only one permutation such that $m(\pi) = n+1$.

Since there are $(n+1)!$ permutations in $S_{n+1}$, one has

$$ (n+1)! = \sum_{m=0}^{n-1} (n-m)(n-m)! + 1 = \sum_{k=1}^n k \cdot k! + 1. $$

Answer 2

Write $k(k!)=((k+1)-1)k!=(k+1)!-k!$. Now your sum telescopes $$ \sum_{k=1}^nk(k!) = \sum_{k=1}^n\Bigl((k+1)!-k!\Bigr)=(n+1)!-1!=(n+1)!-1. $$

Answer 3

$$\sum_{k=1}^nk\cdot k!=\sum_{k=1}^n [(k+1)-1]k!=\sum_{k=1}^n(k+1)!-k!$$ This is a telescoping sum: $$ {2!}-1!+{3!}-{2!}+{4!}-{3!}+\dots+(n+1)!-{n!}= $$ $$ \not{2!}-1!+\not{3!}-\not{2!}+\not{4!}-\not{3!}+\dots+(n+1)!-\not{n!}=(n+1)!-1! $$

Answer 4

By the integral representation of the factorial, this becomes

$$S=\int_0^\infty\sum_{k=0}^nkx^ke^{-x}~\mathrm dx$$

By geometric series and its derivative,

$$S=\int_0^\infty xe^{-x}\frac\partial{\partial x}\frac{1-x^{n+1}}{1-x}~\mathrm dx$$

Integrate by parts,

$$S=\int_0^\infty e^{-x}(x^{n+1}-1)~\mathrm dx$$

And reapply the integral representation of the factorial to finish off with

$$S=(n+1)!-0!$$

Answer 5

Observe that $$ k (k!) = (k+1)! - k! $$ and write the sum as a telescopic sum: $$ \sum_{k=1}^n k (k!) = \sum_{k=1}^n [ (k+1)! - k!] = (n+1)! - 1. $$

Answer 6

Add to the sum $1!+2!+3!+\ldots+n!=\sum _{k=1}^n k!$

You get

$(1+1!)+(2\times2! + 2!) + (3\times 3!+3!)+\ldots +(n\times n!+n!)=\\=2! +2! (2+1)+3!(3+1)+\ldots+n!(n+1)=2!+3!+4!+\ldots+(n+1)!$

Now subtract $1!+2!+3!+\ldots+n!=\sum _{k=1}^n k!$

You have $2!+3!+4!+\ldots+(n+1)! - 1!-2!-3!-\ldots-n!=(n+1)!-1$