I was trying to find the answer to this question
$$ S= \sum_{r=1}^{n}\frac{rn^{r-1}}{\prod_{k=1}^{r}(n+k)} $$
I tried finding a series which S is a derivative of, seeing that the numerator is power rule like, but wasn't able to do so. Solutions or hints would be appreciated.
Let $f_n(x)=x^n/\prod_{k=1}^n(x+k)$ for $n\geqslant 0$ (so that $f_0(x)=1$), then for $r>0$ $$f_{r-1}(x)-f_r(x)=f_{r-1}(x)\left(1-\frac{x}{x+r}\right)=\frac{rx^{r-1}}{\prod_{k=1}^r(x+k)},$$ hence $$\sum_{r=1}^n\frac{rx^{r-1}}{\prod_{k=1}^r(x+k)}=\sum_{r=1}^n\big(f_{r-1}(x)-f_r(x)\big)=f_0(x)-f_n(x)=1-\prod_{k=1}^n\frac{x}{x+k}.$$ Put $x=n$ to obtain $S=1-n^n n!/(2n)!$ already found by others.
Using Pochhammer symbols $$S_n=\sum_{r=1}^{n}\frac{r\,n^{r-1}}{\prod_{k=1}^{r}(n+k)}=\sum_{r=1}^{n}\frac{r\, n^{r-1}}{(n+1)_r}$$ Using Wolfram Alpha and simplifying $$S_n=1-\frac{n^n\sqrt{\pi } }{4^n\,\Gamma \left(n+\frac{1}{2}\right)}$$ which is the same as the result @Parcly Taxel gave in comments (before I answered).
When $n$ becomes large $$S_n\sim 1-\frac 1 {\sqrt 2} e^{-n (2 \log (2)-1)}$$
Edit
If we consider $$f(x)=\sum_{r=1}^{n}\frac{r\,n^{r-1}}{\prod_{k=1}^{r}(n+k)} x^r$$ there is an ugly explicit expression for $f(x)$ but I have not been able to integrate it.
