TAOCP 1.2.6 Exercise 52
Abel's binomial formula (16) states that (x+y)^n = \sum_k \binom{n}{k}x(x-kz)^{k-1}(y+kz)^{n-k}.
Section 1.2.6: Binomial Coefficients
Exercise 52. [HM11] Prove that Abel's binomial formula (16) is not always valid when $n$ is not a nonnegative integer, by evaluating the right-hand side when $n=x=-1$, $y=z=1$.
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Solution
Abel's binomial formula (16) states that
$$ (x+y)^n = \sum_k \binom{n}{k}x(x-kz)^{k-1}(y+kz)^{n-k}. \tag{16} $$
Take
$$ n=-1,\qquad x=-1,\qquad y=1,\qquad z=1. $$
The left-hand side becomes
$$ (x+y)^n = (-1+1)^{-1} = 0^{-1}, $$
which is undefined.
Now evaluate the right-hand side. Since $n=-1$,
$$ \binom{-1}{k} = \frac{(-1)(-2)\cdots(-k)}{k!} = (-1)^k, \qquad k\ge0, $$
by equation (3). Also,
$$ x-kz=-1-k=-(k+1), $$
hence
$$ x(x-kz)^{k-1} = (-1)\bigl(-(k+1)\bigr)^{k-1} = (-1)^k(k+1)^{k-1}. $$
Furthermore,
$$ y+kz=1+k=k+1, $$
so
$$ (y+kz)^{n-k} = (k+1)^{-1-k}. $$
Therefore the $k$th term of the sum equals
$$ \binom{-1}{k}x(x-kz)^{k-1}(y+kz)^{-1-k} = (-1)^k\cdot (-1)^k (k+1)^{k-1}(k+1)^{-1-k}. $$
Since $(-1)^k(-1)^k=1$ and
$$ (k+1)^{k-1}(k+1)^{-1-k} = (k+1)^{-2}, $$
the right-hand side is
$$ \sum_{k\ge0}\frac1{(k+1)^2} = \sum_{m\ge1}\frac1{m^2}. $$
Thus the right-hand side converges to
$$ \sum_{m\ge1}\frac1{m^2} = \frac{\pi^2}{6}. $$
Hence the right-hand side is defined and equal to $\pi^2/6$, while the left-hand side is undefined. Abel's formula therefore fails when $n$ is not a nonnegative integer.
The evaluated right-hand side is
$$ \boxed{\frac{\pi^2}{6}}. $$