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Math Overflow answer https://mathoverflow.net/a/297916/113033 references the binomial identity

\begin{equation} \sum_{t} \binom{r}{t} \frac{(-1)^t}{r+t+1} \binom{r+t+1}{j} =\begin{cases} \frac{1}{(2r+1) \binom{2r}r}, & \text{if } j=0;\\ \frac{(-1)^r}{j} \binom{r}{2r-j+1}, & \text{if } j>0. \end{cases} \end{equation}

However, I did not succeed in search of literature reference of it.

Is there any literature mention of it or proof?

Petro Kolosov
  • 524

1 Answers1

11

Since OP is also asking for a proof ....

We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k}\tag{1} \end{align*} We start with the easy part:

  • $j=0$:

    We obtain \begin{align*} \color{blue}{\sum_{t}}\color{blue}{\binom{r}{t}\frac{(-1)^t}{r+t+1}} &=\int_{0}^1\sum_{t}\binom{r}{t}(-1)^tz^{r+t}\,dz\tag{2.1}\\ &=\int_{0}^1z^r(1-z)^r\,dz\tag{2.2}\\ &\,\,\color{blue}{=\frac{1}{2r+1}\binom{2r}{r}^{-1}}\tag{2.3} \end{align*} and the claim follows.

Comment:

  • In (2.1) we use $\frac{1}{n+1}=\int_{0}^1z^n\,dz$.

  • In (2.2) we factor out $z^r$ and apply the binomial theorem.

  • In (2.3) we we write the reciprocal of a binomial coefficient using the beta function \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz \end{align*}

  • $j>0$:

    We obtain \begin{align*} \color{blue}{\sum_{t}}&\color{blue}{\binom{r}{t}\frac{(-1)^t}{r+t+1}\binom{r+t+1}{j}}\\ &=\frac{1}{j}\sum_{t}\binom{r}{t}\binom{r+t}{j-1}(-1)^t\tag{3.1}\\ &=\frac{1}{j}\sum_{t}\binom{r}{t}[z^{j-1}](1+z)^{r+t}(-1)^t\tag{3.2}\\ &=\frac{1}{j}[z^{j-1}](1+z)^r\sum_{t}\binom{r}{t}(-1)^t(1+z)^t\tag{3.3}\\ &=\frac{1}{j}[z^{j-1}](1+z)^r\left(1-(1+z)\right)^r\tag{3.4}\\ &=\frac{1}{j}[z^{j-1}](1+z)^r(-z)^r\\ &=\frac{(-1)^r}{j}[z^{j-1-r}](1+z)^r\tag{3.5}\\ &=\frac{(-1)^r}{j}\binom{r}{j-1-r}\tag{3.6}\\ &\,\,\color{blue}{=\frac{(-1)^r}{j}\binom{r}{2r-j+1}}\tag{3.7} \end{align*} and the claim follows.

Comment:

  • In (3.1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

  • In (3.2) we apply (1).

  • In (3.3) we factor out terms independent of $t$.

  • In (3.4) we use the binomial theorem.

  • In (3.5) we apply $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

  • In (3.6) we select the coefficient of $z^{j-1-r}$.

  • In (3.7) we use $\binom{p}{q}=\binom{p}{p-q}$.

Markus Scheuer
  • 108,315