<s>[INST] If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function 
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$ 
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?
 [/INST] By using partial fractions, chance upon the solution $A_{0}=1,$ and $A_{1}=-1$. For $n>1$, in particular for all $n=7,$ it seems that
$$A_{7}=(x+7)(\frac{(7)!}{x(x+1)...(x+7)}-\frac{6!}{x(x+1)...(x+6)})
\\=6!\frac{7-(x+7)}{x(x+1)...(x+6)}=6!\frac{-x}{x(x+1)...(x+6)}
\\=-\frac{6!}{(x+1)...(x+6)}.$$ 
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