How to show $\sum_{n=1}^{a+b} \frac{(-1)^n}{n}\binom{n}{n-a,n-b,a+b-n}=0$

Question

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 $i+j+k=n$)

Answer 1

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.