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

Question

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?

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Answer 1

I guess you mean $\displaystyle \sum_{k=0}^n f(k)=n!$

We need $f(0)=0!=1$.

For $n>1$, $\displaystyle f(n)=\displaystyle \sum_{k=0}^n f(k)-\displaystyle \sum_{k=0}^{n-1} f(k)=n!-(n-1)!=(n-1)\cdot(n-1)!$

Answer 2

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.