Factorial Formula Generalisation

Question

How can I generalise the following formula? It seems to be linked somehow to factorial.

$$1 + N + (N-1) N + (N-2) (N-1) N + (N-3) (N-2) (N-1) N \dots$$

Answer 1

$$\sum_{n=0}^\infty \frac{N!}{(N-n)!}$$

Answer 2

Hint: $N=N!/(N-1)!$, $(N-1)N=N!/(N-2)!$ ....

Answer 3

$$ \sum_{j=0}^{\infty}\left[\prod_{k=0}^{j-1}(N-k)\right] $$

In terms of the factorial, this is $$ \sum_{j=0}^{\infty}\frac{N!}{(N-j)!} $$

Answer 4

Small addendum: Note, the expression has only finitely many summands which are not equal to zero, as the terms eventually contain a factor $$(N-N)=0$$

We can write \begin{align*} 1+N+(N-1)N+(N-2)(N-1)N+\cdots&=\sum_{n=0}^{N-1}\frac{N!}{(N-n)!}\tag{1}\\ &=N!\sum_{n=0}^{N-1}\frac{1}{(N-n)!}\tag{2}\\ &=N!\sum_{n=0}^{N-1}\frac{1}{(n+1)!}\tag{3}\\ &=N!\sum_{n=1}^N\frac{1}{n!}\tag{4} \end{align*}

Comment:

• In (1) we can set the upper limit of the sum to $N-1$ since all other summands are $0$.

• In (2) we factor out $N!$ which does not depend on the summation index $n$.

• In (3) we revert the order of summation by letting $n\rightarrow N-1-n$.

• In (4) we shift the index $n$ by one to start with $n=1$.