TITLE: A problem on limit
QUESTION [7 upvotes]: It seems 

$\lim_{n \rightarrow \infty} \sum_{r=1}^{n-1}\binom{n}{r-1}
 \sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}$

is $\frac{1}{2}$.  Here's a plot of the function for $n \leq 300$:

But I can not prove this. Any hints?

REPLY [3 votes]: Note that the $n^\text{th}$ forward difference of $1/x$ is
$$
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac1{k+x}=\frac{n!}{x(x+1)(x+2)\dots(x+n)}\tag{1}
$$
To prove $(1)$, apply the Heaviside Method for Partial Fractions to the right hand side, as is done in this answer with $f(j)=1$.
Use $(1)$ with $n\mapsto n-r+1$, then set $x=r+1$ to get the first step below:
$$
\begin{align}
&\sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1}(-1)^j\binom{n-r+1}{j}\frac1{r+j+1}\\
&=\sum_{r=1}^{n-1}\binom{n}{r-1}
\frac{(n-r+1)!\,r!}{(n+2)!}\\
&=\sum_{r=1}^{n-1}\frac{r}{(n+1)(n+2)}\\
&=\frac12\frac{n(n-1)}{(n+1)(n+2)}\tag{2}
\end{align}
$$
Taking the limit as $n\to\infty$ gives $\frac12$.