Can the factorial function be written as a sum?

Question

I know of the sum of the natural logarithms of the factors of n! , but would like to know if any others exist.

Answer 1

This one is pretty important: $$n! = \sum_{\sigma\in S_n} 1$$

Edit: As Arkamis explains, $S_n$ is the symmetric group on $n$ letters. Each $\sigma\in S_n$ is a permutation on the set $[1,2,\ldots,n]$. Since $S_n$ is a finite set, we may sum a function over it, and the sum of the constant function $f(\sigma)=1$ is just the size of the set, which is $|S_n| = n!$.

Arguably, summing a constant function is cheating. Here's one way to raise the stakes. Let $B_n$ be the set of $n\times n$ integer matrices $A$ such that every sum of a subset of entries from $A$ is in $[0,n]$. Then:

$$n!=\sum_{A\in B_n}|\det A|$$

This is the same identity in a more interesting disguise. Every $n\times n$ permutation matrix $A$ is a member of $B_n$, and $\det A = \pm 1$. On the other hand, if $A\in B_n$ is not a permutation matrix, then you can prove that $\det A = 0$.

Answer 2

Sure, for $n\ge 1$ we have $$n!=\fbox{$(n-1)\times (n-1)!$} + \fbox{$(n-1)!$}$$

Answer 3

If you really want, we can write $n!$ using this infinite series

Answer 4

Lots of good answers, but the answer is quite easily, "yes."

Integer multiplication is repeated addition.

$n!$ is $n$ groups of $(n-1)!$ objects, so $n! = \sum_{k=1}^n (n-1)!$.

Then, $(n-1)!$ is $n-1$ groups of $(n-2)!$ objects, so $n! = \sum_{k_1=1}^n \sum_{k_2 = 1}^{n-1} (n-2)!$, and so on.

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Example:

$$4! = \sum_{k_1 = 1}^4 3! = 3!+3!+3!+3! = 4\cdot 3!.$$ $$\begin{align*} 4! &= \sum_{k_1=1}^4 \sum_{k_2=1}^3 2! \\ &= \sum_{k_1=1}^4 2!+2!+2! \\ &= 2!+2!+2! + 2!+2!+2! + 2!+2!+2! + 2!+2!+2! \\ &= 12\cdot 2! \\ &=4\cdot 3\cdot 2!. \end{align*}$$

Answer 5

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

Answer 6

$n! = e^{ \sum_{k = 1}^n \ln (k)}$

Answer 7

$$n!=\prod_{p<n}p^{\sum_{k=1}^\infty\lfloor{n\over p^k}\rfloor}$$

Answer 8

Another combinatorially important sum: $\displaystyle n! = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$, where $\displaystyle\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ is the (unsigned) Stirling symbol of the first kind, which can be defined (for instance) by the relation $x(x+1)\ldots(x+n-1) = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr] x^k$. (And of course, by setting $x=1$ in this relation we get the initial sum.) The combinatorial significance is that $\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ counts the number of permutations of the numbers $1..n$ with $k$ distinct cycles; the relation $\displaystyle n! = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ thus says that the total number of permutations is just the sum over all $k$ of the number of permutations into $k$ cycles.

Answer 9

I have found the following formula: $$n!=\sum_{k=2}^{n+1}(-1)^{n+1-k}\binom{n+1}{k}\sum_{i=1}^{k-1}i^{n},\ n\in N.$$

Answer 10

$$\frac{(n+2)!}{2} = (n+1)! + \sum_{i=1}^n i\,n!$$