A problem on limit

Question

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?

Answer 1

$\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.

Answer 2

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 $\lim_{n\to\infty}\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}(-1)^{j}\binom{n-r+1}{j}\big(\frac{1}{r+j+1}\big)=\frac{1}{2}$.

Answer 3

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 $\frac12$.