A Curious Binomial Sum Identity without Calculus of Finite Differences

Let $f$ be a polynomial of degree $m$ in $t$. The following curious identity holds for $n \geq m$,
\begin{align}
\binom{t}{n+1} \sum_{j = 0}^{n} (-1)^{j} \binom{n}{j} \frac{f(j)}{t – j} = (-1)^{n} \frac{f(t)}{n + 1}.
\end{align}
The proof follows by transforming it into the identity
\begin{align}
\sum_{j = 0}^{n} \sum_{k = j}^{n} (-1)^{k-j} \binom{k}{j} \binom{t}{k} f(j) = \sum_{k = 0}^{n} \binom{t}{k} (\Delta^{k} f)(0) = f(t),
\end{align}
where $\Delta^{k}$ is the $k^{\text{th}}$ forward difference operator. However, I’d like to prove the aforementioned identity directly, without recourse to the calculus of finite differences. Any hints are appreciated!

Thanks.

Solutions Collecting From Web of "A Curious Binomial Sum Identity without Calculus of Finite Differences"

This is just Lagrange interpolation for the values $0, 1, \dots, n$.

This means that after cancelling the denominators on the left you can easily check that the equality holds for $t=0, \dots, n$.

I just ran across this question after working on this answer, and realized that the same method could be used here.

Notice that your equation is equivalent to
$$
\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}\frac{f(j)}{t-j}=\frac{n!f(t)}{t(t-1)(t-2)\dots(t-n)}
$$
As long as $f$ is a polynomial of degree $n$ or less, apply the Heaviside Method for Partial Fractions to the right hand side to get the left hand side.

That is, to compute the coefficient of $\frac1{t-j}$ on the left hand side, multiply both sides by $t-j$ and set $t=j$. The right hand side becomes
$$
\frac{n!f(j)}{j(j-1)(j-2)\dots1(-1)(-2)\dots(j-n)}=(-1)^{n-j}\binom{n}{j}f(j)
$$

This can also be done with complex variables. Observe that we have by
inspection that
$$\sum_{j=0}^n (-1)^j {n\choose j} \frac{f(j)}{t-j}
= (-1)^n \times n! \times
\sum_{k=0}^n \mathrm{Res}_{z=k}
\frac{f(z)}{t-z}\prod_{q=0}^n \frac{1}{z-q}.$$

This holds even if $f(t)$ vanishes at some positive integer in the
range.

Recall that the residues sum to zero, so the right is equal to
$$-(-1)^n \times n! \times
\sum_{k\in\{\infty,t\}} \mathrm{Res}_{z=k}
\frac{f(z)}{t-z}\prod_{q=0}^n \frac{1}{z-q}.$$

The residue at infinity of a function $h(z)$ is given by the formula
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0}
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$

which in the present case yields
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{f(1/z)}{t-1/z}\prod_{q=0}^n \frac{1}{1/z-q}
\\ = -\mathrm{Res}_{z=0} \frac{z^{n+1}}{z^2}
\frac{f(1/z)}{t-1/z}\prod_{q=0}^n \frac{1}{1-qz}
\\ = -\mathrm{Res}_{z=0} z^n
\frac{f(1/z)}{zt-1}\prod_{q=0}^n \frac{1}{1-qz}.$$

Note however that $f(z)$ has degree $m$ and we require that $n\ge m$
which means $f(1/z) z^n$ has no pole at zero and hence the residue at
infinity is zero as well.

That leaves the residue at $z=t$ for a total contribution of
$$-(-1)^n \times n! \times (-1)\times f(t)\times
\prod_{q=0}^n \frac{1}{t-q}
= (-1)^n f(t) \times n! \prod_{q=0}^n \frac{1}{t-q}
\\ = (-1)^n f(t) \times n! \times
\frac{1}{(n+1)!} {t\choose n+1}^{-1}
\\ = (-1)^n \frac{f(t)}{n+1} {t\choose n+1}^{-1}$$
which was to be shown, QED.