Is there a sum of “n+j” terms equivalent to n!

Is there any function so that...

$$\sum_{k=0}^{n} f(k) = n!$$

or,

$$\sum_{k=0}^{n+j} f(k) = n!$$

where j is any arbitrary integer?

• $$n! + \underbrace{0 + \ldots + 0}_{j - 1 \text{ times}}?$$ – Theo Bendit May 23 at 2:40
• @FelixMarin Oh yeah, my bad. – FamiliarTheory May 23 at 8:24
• @TheoBendit $\displaystyle\Huge\left(\bullet\qquad\bullet \atop {\mid \atop \smile}\right)$. Quite right. – Felix Marin May 23 at 16:06

I guess you mean $$\displaystyle \sum_{k=0}^n f(k)=n!$$
We need $$f(0)=0!=1$$.
For $$n>1$$, $$\displaystyle f(n)=\displaystyle \sum_{k=0}^n f(k)-\displaystyle \sum_{k=0}^{n-1} f(k)=n!-(n-1)!=(n-1)\cdot(n-1)!$$
From the tag 'polynomial', I guess what you mean might be that if there exists a polynomial $$f(x)$$ such that $$\sum_{k=0}^n f(k)=n!.$$ If so, then we could induce that $$f(n)=n!-(n-1)!$$ for $$n>0$$. Since $$f$$ is differentiable, so for any $$n>0$$ there exists a point $$x_n\in [n,n+1]$$ such that $$f'(x_n)=\frac{f(n+1)-f(n)}{(n+1)-n}=(n+1)!-(n-1)!=(n^2+n-1)(n-1)!.$$ Note that $$f'(x)=a_tx^{t-1}+\cdots+a_1$$ where $$t=$$deg$$f$$, hence $$\dfrac{f'(x_n)}{x_n^{t-1}}\rightarrow a_t$$ as $$n\rightarrow \infty$$. However $$\dfrac{f'(x_n)}{x_n^{t-1}}=\dfrac{(n^2+n-1)(n-1)!}{x_n^{t-1}}\geq \dfrac{(n^2+n-1)(n-1)!}{(n+1)^{t-1}}\rightarrow +\infty$$ which contradicts. So the polynomial satisfying the condition in the question does not exist.