<s>[INST] 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 $i+j+k=n$)
 [/INST] 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.
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