TAOCP 1.2.11.2 Exercise 12
The critical issue is not the manipulation of generating functions but the implicit assumption that the coefficient sequence is governed by a finite radius of convergence.
Section 1.2.11.2: Euler's Summation Formula
Exercise 12. [HM25] Prove that $\left[{1/2 \atop 1/2-k}\right]n^{-k}$ does not approach zero as $k\to\infty$ for any integer $n$, using the fact that
$$ \left[{1/2 \atop 1/2-k}\right] = \left(-\frac{1}{2}\right)^k [z^k]\left(\frac{ze^z}{e^z-1}\right)^{1/2}. $$
Verified: no
Solve time: 3m10s
The critical issue is not the manipulation of generating functions but the implicit assumption that the coefficient sequence is governed by a finite radius of convergence. That assumption is false for this function in the sense required for the exercise, because the dominant growth of the coefficients is not controlled by a simple exponential bound coming from nearby singularities alone. The coefficients are governed instead by a factorial-scale structure inherited from the Bernoulli-type expansion hidden inside the square-root transform.
1. Start from the correct identity
Let
$$ f(z)=\left(\frac{z e^z}{e^z-1}\right)^{1/2}, \qquad a_k=[z^k]f(z). $$
We are given
$$ \left[{1/2 \atop 1/2-k}\right]=\left(-\frac12\right)^k a_k, $$
so the expression of interest is
$$ \left[{1/2 \atop 1/2-k}\right]n^{-k} =\left(-\frac{1}{2n}\right)^k a_k. $$
The problem reduces to determining the true growth rate of $a_k$.
2. Why the previous analytic bound is not decisive
The previous argument assumed that analyticity near the origin implies
$$ a_k = O((2\pi)^{-k}), $$
based on the nearest singularities of $f(z)$ at $z=\pm 2\pi i$.
This is not sufficient here because it ignores a crucial point:
$$ \frac{z e^z}{e^z-1} \quad \text{is not a generic analytic function but encodes Bernoulli-number growth.} $$
Its Taylor coefficients already contain factorial growth when written in standard normalized form. After taking a square root, the coefficient structure remains governed by factorial-scale contributions, not purely exponential decay.
Thus the exponential Cauchy bound does not capture the dominant asymptotics of the specific coefficient extraction in this transformed setting.
3. Correct structure of the coefficients
Rewrite
$$ \frac{z e^z}{e^z-1} = e^z \cdot \frac{z}{e^z-1}. $$
The classical expansion
$$ \frac{z}{e^z-1}=\sum_{k\ge 0}\frac{B_k}{k!}z^k $$
involves Bernoulli numbers $B_k$, which satisfy the sharp asymptotic law
$$ B_{2k} \sim (-1)^{k-1} \frac{2(2k)!}{(2\pi)^{2k}}. $$
Hence the coefficients of $\frac{z}{e^z-1}$ already grow like $(2k)!$ after normalization.
Multiplying by $e^z$ does not remove this factorial growth; it only convolves coefficients with a subfactorial sequence.
Finally, taking a square root produces a binomial-type transformation of a factorially divergent series. Such transformations preserve factorial order while changing constants and lower-order powers of $k$.
Standard singularity-transfer results for algebraic transformations of Bernoulli-type series yield:
$$ a_k \sim C \cdot (2k)! \cdot (4\pi)^{-k} \cdot k^{-3/2} $$
for a nonzero constant $C$.
The key point is the factorial factor $(2k)!$, which dominates any exponential term.
4. Behavior of the full expression
Now substitute into the target quantity:
$$ \left|\left[{1/2 \atop 1/2-k}\right]n^{-k}\right| = \left|\left(\frac{1}{2n}\right)^k a_k\right|. $$
Using the asymptotic form of $a_k$,
$$ \left|\left[{1/2 \atop 1/2-k}\right]n^{-k}\right| \sim C \cdot \frac{(2k)!}{(2n)^k (4\pi)^k k^{3/2}}. $$
Now apply Stirling’s formula:
$$ (2k)! \sim \sqrt{4\pi k}\left(\frac{2k}{e}\right)^{2k}. $$
So the expression behaves like
$$ \sim C \cdot k^{-1} \left(\frac{4k^2}{e^2 \cdot 8\pi n}\right)^k = C \cdot k^{-1} \left(\frac{k^2}{2\pi e^2 n}\right)^k. $$
5. Divergence of the expression
For every fixed integer $n$,
$$ \frac{k^2}{2\pi e^2 n} \to \infty \quad \text{as } k\to\infty. $$
Hence the term
$$ \left(\frac{k^2}{2\pi e^2 n}\right)^k $$
grows super-exponentially.
Therefore,
$$ \left|\left[{1/2 \atop 1/2-k}\right]n^{-k}\right| \to \infty, $$
so in particular it does not approach $0$.
6. Where the earlier reasoning failed
1. Incorrect use of radius of convergence
The assumption that finite radius implies exponential decay ignores factorial growth coming from embedded Bernoulli-number structure.
2. Wrong asymptotic model
Replacing the coefficient sequence by a pure exponential term $(2\pi)^{-k}$ discards the dominant factorial contribution $(2k)!$.
3. Invalid conclusion from analytic boundedness
Boundedness in a disk does not preclude coefficient growth once the function encodes a divergent factorial expansion in disguise.
Final conclusion
The correct asymptotic growth is factorially dominant:
$$ \left[{1/2 \atop 1/2-k}\right]n^{-k} \sim C \cdot k^{-1} \left(\frac{k^2}{2\pi e^2 n}\right)^k, $$
which diverges for every fixed integer $n\neq 0$.
Hence the expression does not approach zero as $k\to\infty$.