Sum of products of binomial coefficients divided by the index I am having trouble in finding a more compact form for the following series.
$$\sum_{i=j}^{n}\left(\begin{array}{c}n \\ i\end{array}\right)\left(\begin{array}{c}i \\ j\end{array}\right) \frac{(-1)^{i+j}}{i}, \quad n\ge 1, j\le n-1.$$
$j$ can be regarded as a constant. I think I saw something similar to this before but I am not sure if it is possible to simplify the sum further. Any suggestions are greatly appreciated.
 A: Note that
$$j\sum_{i=j}^n\binom{n}i\binom{i}j\frac{(-1)^{i+j}}i=\sum_{i=j}^n(-1)^{i-j}\binom{n}i\binom{i-1}{j-1}\,,$$
so it suffices to show that
$$\sum_{i=j}^n(-1)^{i-j}\binom{n}i\binom{i-1}{j-1}=1\,.\tag{1}$$
For $j\le i\le n$ there are $\binom{n}i\binom{i-1}{j-1}$ pairs $\langle S,T\rangle$ of subsets of $[n]$ such that $T\subseteq S$, $|S|=i$, $|T|=j$, and $\max T=\max S$. How many net times does a given $T$ get counted on the lefthand side of $(1)$?
Let $T$ be a $j$-subset of $[n]$ with $\max T=m\ge j$. For $i=j,\ldots,m$ there are $\binom{m-j}{i-j}$ $i$-subsets $S$ of $[n]$ such that $m=\max S$, so $T$ gets counted
$$\sum_{i=j}^m(-1)^{i-j}\binom{m-j}{i-j}=\sum_{\ell=0}^{m-j}(-1)^\ell\binom{m-j}\ell=\begin{cases}
0,&\text{if }m>j\\
1,&\text{if }m=j
\end{cases}$$
times, and $(1)$ follows immediately.
A: We want to show that
$$\sum_{i=j}^n\binom{n}{i}\binom{i}{j} \frac{(-1)^{i+j}}{i} = \frac{1}{j},$$
equivalently,
$$\sum_{i=j}^n\binom{n}{i}\binom{i-1}{j-1} (-1)^{i+j} = 1. \tag1$$
Apply snake oil:
\begin{align}
&\sum_{j=1}^n \sum_{i=j}^n\binom{n}{i}\binom{i-1}{j-1} (-1)^{i+j} z^j \\
&= \sum_{i=1}^n \binom{n}{i}(-1)^i \sum_{j=1}^i\binom{i-1}{j-1} (-z)^j \\
&= \sum_{i=1}^n \binom{n}{i}(-1)^i (-z)(1-z)^{i-1} &&\text{(binomial theorem)}\\
&= \frac{z}{1-z}\sum_{i=1}^n \binom{n}{i} (1-z)^i &&\\
&= \frac{z}{1-z} (z^n-1) &&\text{(binomial theorem)}\\
&= \sum_{j=1}^n z^j&&\text{(finite geometric series)}
\end{align}
which immediately implies $(1)$ by extracting the coefficient of $z^j$.
A: Supposing that we are interested in
$$S_{n, q} =
\sum_{p=q}^n {n\choose p} {p\choose q} \frac{(-1)^{p+q}}{p}$$
we observe that
$${n\choose p} {p\choose q} =
\frac{n!}{(n-p)! \times q! \times (p-q)!}
= {n\choose q} {n-q\choose n-p}$$
so we obtain
$${n\choose q} \sum_{p=q}^n {n-q\choose n-p}
\frac{(-1)^{p+q}}{p}
= {n\choose q} \sum_{p=0}^{n-q} {n-q\choose n-q-p}
\frac{(-1)^{p}}{p+q}
\\ = {n\choose q} \sum_{p=0}^{n-q} {n-q\choose p}
\frac{(-1)^{n-q-p}}{n-p}.$$
We seek to evaluate the intermediate sum
$$T_{n,q} = \sum_{p=0}^{n-q} {n-q\choose p}
\frac{(-1)^{p}}{n-p}.$$
With this in mind we introduce
$$f(z) = \frac{(n-q)! \times (-1)^{n-q}}{n-z}
\prod_{r=0}^{n-q} \frac{1}{z-r}.$$
This has the property that for $0\le p\le n-q$ we have
$$\mathrm{Res}_{z=p} f(z) =
\frac{(n-q)! \times (-1)^{n-q}}{n-p}
\prod_{r=0}^{p-1} \frac{1}{p-r}
\prod_{r=p+1}^{n-q} \frac{1}{p-r}
\\ = \frac{(n-q)! \times (-1)^{n-q}}{n-p}
\frac{1}{p!} \frac{(-1)^{n-q-p}}{(n-q-p)!}
= {n-q\choose p} \frac{(-1)^p}{n-p}$$
so that
$$T_{n,q} = \sum_{p=0}^{n-q} \mathrm{Res}_{z=p} f(z).$$
Now residues sum to zero and the residue at infinity is zero by
inspection hence
$$T_{n,q} = - \mathrm{Res}_{z=n} f(z) =
(n-q)! (-1)^{n-q} \prod_{r=0}^{n-q} \frac{1}{n-r}
\\ = (n-q)! (-1)^{n-q} \frac{(q-1)!}{n!}
= (-1)^{n-q} \frac{1}{q} {n\choose q}^{-1}.$$
We thus obtain
$$S_{n,q} =
{n\choose q} (-1)^{n-q} (-1)^{n-q} {n\choose q}^{-1} \frac{1}{q}.$$
This simplifies to
$$\bbox[5px,border:2px solid #00A000]{
S_{n,q} = \frac{1}{q}.}$$
A: I will just post another feasible method to prove this identity.
It is straightforward to show that
$$ {{i+j}\choose{j}}{{n}\choose{i+j}}= {{n}\choose{j}}{{n-j}\choose{i}}. \tag1 $$
By the definition of the Hypergeometric function, we have
$$ _2F_1(j-n,j;j+1;z) =\sum_{i=0}^{n-j}(-1)^i {{n-j}\choose{i}} \frac{j}{j+i}z^i\quad \tag2$$
since $j<n$.
Combining (1) and (2), it is immediate that
$$ \sum_{i=j}^n (-z)^i {{i}\choose{j}}{{n}\choose{i}}\frac{1}{i} = \frac{(-z)^j}{j}{{n}\choose{j}}{}_2F_1(j-n,j;j+1;z).$$
Let $z=1$, then by Gauss's theorem, one can simplify the Hypergeometric function and get the same result as shown in other solutions.
