Sum of $n$ terms of the given series. Find the sum of $n$ terms of the series:
$$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\frac{4x^3}{(x+1)(x+2)(x+3)(x+4)}+.......$$
Could someone give me slight hint to proceed in this question?
 A: Hint. One may observe that
$$
\frac{nx^{n-1}}{\prod _{k=1}^n (x+k)}=\left(1-\frac{x^n \Gamma(x+1)}{\Gamma(x+n+1)}\right)-\left(1-\frac{x^{n-1} \Gamma(x+1)}{\Gamma(x+n)}\right)
$$ giving
$$
\sum_{n=1}^N\frac{nx^{n-1}}{\prod _{k=1}^n (x+k)}=1-\frac{x^N \Gamma(x+1)}{\Gamma(x+N+1)}
$$
A: Recalling that the lower incomplete gamma function has the power series $$\gamma\left(s,x\right)=x^{s-1}\Gamma\left(s\right)e^{-x}\sum_{k\geq1}\frac{x^{k}}{\Gamma\left(s+k\right)}\tag{1}$$ we have $$\frac{d}{dx}\left(\frac{\gamma\left(s,x\right)}{x^{s-1}\Gamma\left(s\right)e^{-x}}\right)=\sum_{k\geq1}\frac{kx^{k-1}}{\Gamma\left(s+k\right)}
 $$ hence $$\sum_{k\geq1}\frac{kx^{k-1}}{\Gamma\left(s+k\right)}=\frac{1+x^{-s}e^{x}\left(x-1-s\right)\gamma\left(s,x\right)}{\Gamma\left(s\right)}$$ so taking $s=x+1$ we get $$\sum_{k\geq1}\frac{kx^{k-1}}{\Gamma\left(x+1+k\right)}=\frac{1}{\Gamma\left(x+1\right)}\tag{2}
 $$ and now recalling that the Pochhammer symbol can be written as $$\left(x+1\right)\cdots\left(x+n\right)=\left(x+1\right)_{n}=\frac{\Gamma\left(x+1+n\right)}{\Gamma\left(x+1\right)}\tag{3}$$ we get $$\sum_{k\geq1}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}=1.
 $$ Then $$\sum_{k=1}^{N}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}=\sum_{k\geq1}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}-\sum_{k=N+1}^{\infty}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}
 $$ $$=1-x^{N}\sum_{k\geq1}\frac{\left(k+N\right)x^{k-1}}{\left(x+1\right)\cdots\left(x+k+N\right)}=1-\frac{x^{N}\Gamma\left(x+1\right)}{\Gamma\left(x+N+1\right)}$$ using $(1)$, $(2)$ and $(3)$.
