How to show $\sum_{n=1}^{a+b} \frac{(-1)^n}{n}\binom{n}{n-a,n-b,a+b-n}=0$ I was trying to prove directly that, for $a,b$ positive integers:
$$\sum_{n=1}^{a+b} \frac{(-1)^n}{n}\binom{n}{n-a,n-b,a+b-n}=0$$
Alternatively:
$$\sum_{n=1}^{a+b}\frac{(-1)^n}{n}\binom{n}{a}\binom{a}{n-b}=0$$
This comes from trying to prove the formal power series identity:

If $f(x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$ then $$f(x)+f(y)=f(x+y-xy)$$

which we know is true because $f(x)=-\log(1-x)$ in the complex numbers when $|x|<1$.
I would like to prove this directly, without referencing logarithm, and it reduces to the above identity.
(Note: $\binom{n}{i,j,k}=\binom{n}{i}\binom{n-i}{j}=\frac{n!}{i!j!k!}$ is the trinomial coefficient, defined when $i+j+k=n$)
 A: It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\sum_{n=1}^{a+b}&\frac{(-1)^n}{n}\binom{n}{a}\binom{a}{n-b}\\
&=\frac{1}{a}\sum_{n=1}^{a+b}(-1)^n\binom{n-1}{a-1}\binom{a}{n-b}\tag{1}\\
&=\frac{1}{a}\sum_{n=0}^{a+b-1}(-1)^{n+1}\binom{n}{a-1}\binom{a}{n+1-b}\tag{2}\\
&=\frac{1}{a}\sum_{n=0}^\infty(-1)^{n+1}[z^{a-1}](1+z)^n[u^{n+1-b}](1+u)^a\tag{3}\\
&=-\frac{1}{a}[z^{a-1}]\sum_{n=0}^\infty(-1)^n(1+z)^n[u^n]u^{b-1}(1+u)^a\tag{4}\\
&=-\frac{1}{a}[z^{a-1}](-(1+z))^{b-1}(1-(1+z))^a\tag{5}\\
&=(-1)^{a+b}\frac{1}{a}[z^{a-1}](1+z)^{b-1}z^a\tag{6}\\
&=0
\end{align*}
  Comment:



*

*In (1) we use the binomial identity
\begin{align*}
\binom{n}{a}=\frac{n}{a}\binom{n-1}{a-1}
\end{align*}

*In (2) we shift the index $n$ by $1$.

*In (3) we apply the coefficient of operator twice. We also extend the limits of the series without changing anything since we are adding zeros only.

*In (4) we use the linearity of the coefficient of operator.

*In (5) we use the substitution rule of the coefficient of operator with $u:= -(1+z)$
\begin{align*}
A(z)=\sum_{n=0}^\infty a_n z^n=\sum_{n=0}^\infty z^n [u^n]A(u)
\end{align*}

*In (6) we observe the coefficient of $z^{a-1}$ is zero.
A: Finally found a combinatorial answer to my own question, six years later.
Letting $i=a+b-n,$ the second form becomes, up to sign:
$$\sum_{i=0}^{a+b-1}(-1)^i\frac1{a+b-i}\binom{a+b-i}{a}\binom{a}{a-i}=0$$
But $$\binom{a}{a-i}=\binom ai$$ and $$\frac{1}{a+b-i}\binom{a+b-i}{a}=\frac1{a}\binom{a+b-i-1}{a-1}=\frac1{a}\binom{a+b-1-i}{b-i}$$
Also, $\binom ai$ is zero if $i>a,$ so we want:
$$\sum_{i=0}^{a}(-1)^i\binom ai\binom{a+b-1-i}{b-i}=0\tag1$$
Now that looks looks like an inclusion-exclusion formula.
Indeed, if $A$ is the set of all subsets of $\{1,2,\dots,a+b-1\}$ of size $b,$ the expression in $(1)$ is the inclusion-exclusion count of all elements of $A$ which do not include $1,2,3,\dots,a,$ which is obviously $0.$
Indeed, we get a more general combinatorial result:

If $m\leq n$ and $k\geq 0,$ then: $$\sum_{i=0}^m(-1)^i\binom{m}i\binom{n-i}{k-i}=\binom{n-m}{k}$$

This can be seen as an inclusion-exclusion count for the case when $M\subseteq N$ With $|M|=m, |N|=n,$ and we are counting all subsets of $N$ of size $k$ which don't contain any element of $M,$ which is just the subsets of size $k$ of $N\setminus M.$
The case $m=a, n=a+b-1, k=b$ gives $(1)$ when $b>0.$
