Evaluate the sum to n terms : $\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\dots$ The main question is :

Evaluate the sum to n terms : 
  $$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\dots$$

My approach :
My first intuition was to make the series telescopic. Following is what I get as the general term.
Let $T_{r}$ denote the $r^{th}$ term.
Therefore, 
$$T_{r} = \frac{r.x^{r-1}}{(x+1)(x+2)\dots (x+r)}$$
I've spent a lot of time using different techniques to manipulate this term by which I can get a telescopic series, but to no avail. There may be different methods to solve this question, but I am interested only in approach involving telescopic series. I am a high school student so I may not understand a few advanced concepts. Please bear that in mind while providing hints or answers. Thanks a lot.
 A: Notice
$$\begin{align}\frac{rx^{r-1}}{\prod_{k=1}^r(x+k)}
&= \frac{rx^r}{\prod_{k=0}^r(x+k)}
= \frac{((x+r)-x)x^r}{\prod_{k=0}^r(x+k)}\\
&= \frac{x^r}{\prod_{k=0}^{r-1}(x+k)} - \frac{x^r}{\prod_{k=1}^r(x+k)}
= \frac{x^{r-1}}{\prod_{k=1}^{r-1}(x+k)} - \frac{x^r}{\prod_{k=1}^r(x+k)}
\end{align}
$$
has the form of a telescoping sum, we have
$$\sum_{r=1}^n \frac{rx^{r-1}}{\prod_{k=1}^r(x+k)}
= 1 - \frac{x^n}{\prod_{k=1}^n(x+k)}$$
A: Consider the following similar problem
$$f_N(x,n)=\sum_{r=1}^N\frac{rx^{r-1}}{(n+1)(n+2)(n+3)\dots(n+r)}=\sum_{r=1}^N\frac{rx^{r-1}n!}{(n+r)!}$$
Integrate $f_N(x,n)$ with respect to $x$:
$$\int f_N(x,n)dx=\int\sum_{r=1}^N\frac{rx^{r-1}n!}{(n+r)!}dx$$
$$=c+\sum_{r=1}^N\frac{x^rn!}{(n+r)!}$$
$$=c+n!e^xx^{-n}\left(\frac{\Gamma(n+N+1,x)}{\Gamma(n+N+1)}-\frac{\Gamma(n+1,x)}{\Gamma(n+1)}\right)$$
Thanks to WolframAlpha, where $\Gamma(x,y)$ is the incomplete gamma function and $\Gamma(x)$ is the gamma function.
Thus, we have
$$f_N(x,n)=\frac{d}{dx}n!e^xx^{-n}\left(\frac{\Gamma(n+N+1,x)}{\Gamma(n+N+1)}-\frac{\Gamma(n+1,x)}{\Gamma(n+1)}\right)$$
And your original problem:
$$\sum_{r=1}^NT_r=f_N(x,x)$$
