A problem on limit 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? 
 A: $\begin{align}
 S(n)
&=\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}\\
&= \sum_{r=0}^{n-2}\binom{n}{r}
 \sum_{j=0}^{n-r} (-1)^j \binom{n-r}{j} \frac{1}{r+j+2}\\
\text{(replace }r \text{ by } n-r)\quad&= \sum_{r=2}^{n}\binom{n}{r}
 \sum_{j=0}^{r} (-1)^j \binom{r}{j} \frac{1}{n-r+j+2}\\
(D \text{ described below) }\quad&= D+\sum_{r=0}^{n}\binom{n}{r}
 \sum_{j=0}^{r} (-1)^j \binom{r}{j} \frac{1}{n-r+j+2}\\
\text{(replace } j \text{ by } r-j)\quad&= D+\sum_{r=0}^{n}\binom{n}{r}
 \sum_{j=0}^{r} (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}\\
\text{(change order of summation) }\quad&= D+\sum_{j=0}^{n}\sum_{r=j}^{n}\binom{n}{r}
  (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}\\
\quad&=D+ \sum_{j=0}^{n}\frac{(-1)^{j}}{n-j+2}\sum_{r=j}^{n}
(-1)^{r}\binom{n}{r}\binom{r}{j} \\
\text{ (put } n-r \text{ for } r)\quad&=D+ \sum_{j=0}^{n}\frac{(-1)^{j}}{n-j+2}
\sum_{r=0}^{n-j}(-1)^{n-r}\binom{n}{n-r}\binom{n-r}{j} \\
\text{ (put } n-j \text{ for } j)\quad&= D+\sum_{j=0}^{n}\frac{(-1)^{n-j}}{j+2}
\sum_{r=0}^{j}(-1)^{n-r}\binom{n}{n-r}\binom{n-r}{n-j} \\
&= D+\sum_{j=0}^{n}\frac{(-1)^{j}}{j+2}
\sum_{r=0}^{j}(-1)^{r}\frac{n!(n-r)!}{(n-r)!r!(n-j)!(j-r)!} \\
&= D+n!\sum_{j=0}^{n}\frac{(-1)^{j}}{(n-j)!(j+2)}
\sum_{r=0}^{j}(-1)^{r}\frac{1}{r!(j-r)!} \\
&= D+n!\sum_{j=0}^{n}\frac{(-1)^{j}}{j!(n-j)!(j+2)}
\sum_{r=0}^{j}(-1)^{r}\frac{j!}{r!(j-r)!} \\
&= D+\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^{j}}{j+2}
\sum_{r=0}^{j}(-1)^{r}\binom{j}{r} \\
\text{(since }\sum_{r=0}^{j}(-1)^{r}\binom{j}{r}=0 \text{ for }j > 0)\quad
&= D+\frac{1}{2}\\
\end{align}
$
$D=
-\sum_{r=0}^1\binom{n}{r}\sum_{j=0}^{r} (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}
$.
If $r=0$,
$\binom{n}{r}\sum_{j=0}^{r}(-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}
=\binom{n}{0}\binom{0}{0} \frac{1}{n+2}
=\frac{1}{n+2}
$.
If $r=1$,
$\binom{n}{r}\sum_{j=0}^{r} (-1)^{r-j} \binom{r}{j} \frac{1}{n-j+2}
=\binom{n}{1}\left(-\binom{1}{0} \frac{1}{n+2}+\binom{1}{1} \frac{1}{n+1}\right)
=n\left(-\frac{1}{n+2}+\frac{1}{n+1}\right)
=n\left(\frac{1}{(n+1)(n+2)}\right)
=\frac{n}{(n+1)(n+2)}
$.
So $D
=-\frac{1}{n+2}
-\frac{n}{(n+1)(n+2)}
=\frac{-(n+1)-n}{(n+1)(n+2)}
=\frac{-2n-1}{(n+1)(n+2)}
=-\frac{2n+1}{(n+1)(n+2)}
$.
Therefore
$S(n) = \frac12-\frac{2n+1}{(n+1)(n+2)}
$.
OMG!!!!!
I can't believe that
all this algebra
came up with something that
looks correct!
Note:
to compare my remainder
with user71352's,
$\frac{2n+1}{(n+1)(n+2)}-\frac{1}{n+1}
=\frac{2n+1-(n+2)}{(n+1)(n+2)}
=\frac{n-1}{(n+1)(n+2)}
$.
A computation could decide which is correct,
but, considering the relative complexity of the solutions,
mine has a higher chance of being wrong.
(Added later)
But it wasn't.
Surprisingly (to me),
mine was correct.
Feels good.
I knew it would.
A: Note that $\int_{0}^{1}x^{r+j}dx=\frac{1}{r+j+1}$.
So the sum becomes:
$\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-1)^{j}\int_{0}^{1}x^{r+j}dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-1)^{j}x^{r+j}dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}x^{r}\big(\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-x)^{j}\big)dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}x^{r}(1-x)^{n-r+1}dx$
$=\int_{0}^{1}\sum_{r=0}^{n-2}\binom{n}{r}x^{r+1}(1-x)^{n-r}dx$
$=\int_{0}^{1}x\sum_{r=0}^{n-2}\binom{n}{r}x^{r}(1-x)^{n-r}dx=\int_{0}^{1}x\big((x+(1-x))^{n}-nx^{n-1}(1-x)-x^{n}\big)dx$
$=\int_{0}^{1}x\big(1-nx^{n-1}+nx^{n}-x^{n}\big)dx=\int_{0}^{1}xdx-n\int_{0}^{1}x^{n}dx+n\int_{0}^{1}x^{n+1}dx-\int_{0}^{1}x^{n+1}dx$
$=\frac{1}{2}-\frac{n}{n+1}+\frac{n}{n+2}-\frac{1}{n+2}=\frac{1}{2}+\frac{-n}{(n+1)(n+2)}-\frac{1}{n+2}=\frac{1}{2}-\frac{2n+1}{(n+1)(n+2)}$
$=\frac{1}{2}-\frac{2+\frac{1}{n}}{1+\frac{1}{n}}\cdot\frac{1}{n+2}$.
So $\lim_{n\to\infty}\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}(-1)^{j}\binom{n-r+1}{j}\big(\frac{1}{r+j+1}\big)=\frac{1}{2}$.
A: 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 $\frac12$.
