TAOCP 7.2.2.2 Exercise 197
The reviewer’s objection reveals that the stated exercise is not correct as written.
Section 7.2.2.2: Satisfiability
Exercise 197. [HM21] Prove that the quotient $q(a, b, A, B, n) = \binom{(a+b)n}{an} \binom{(A+B)n}{An} / \binom{(a+b+A+B)n}{(a+A)n}$ is $\Theta(n^{-1/2})$ as $n \to \infty$, if $a, b, A, B > 0$.
Verified: yes
Solve time: 2m54s
The reviewer’s objection reveals that the stated exercise is not correct as written. The correct Stirling analysis gives an additional exponential factor. The following is the corrected argument.
Edit
Solution
Let
[ q(a,b,A,B,n)= \frac{\binom{(a+b)n}{an}\binom{(A+B)n}{An}} {\binom{(a+b+A+B)n}{(a+A)n}} . ]
Applying Stirling's formula
[ m!\sim \sqrt{2\pi m}\left(\frac me\right)^m ]
to a general binomial coefficient gives, for fixed positive (u,v),
[ \binom{(u+v)n}{un} \sim \frac{1}{\sqrt{2\pi n}} \sqrt{\frac{u+v}{uv}} \left( \frac{(u+v)^{u+v}}{u^u v^v} \right)^n . ]
Therefore,
[ \binom{(a+b)n}{an} \sim \frac{1}{\sqrt{2\pi n}} \sqrt{\frac{a+b}{ab}} \left( \frac{(a+b)^{a+b}}{a^a b^b} \right)^n , ]
and similarly,
[ \binom{(A+B)n}{An} \sim \frac{1}{\sqrt{2\pi n}} \sqrt{\frac{A+B}{AB}} \left( \frac{(A+B)^{A+B}}{A^A B^B} \right)^n . ]
For the denominator,
[ \binom{(a+b+A+B)n}{(a+A)n} \sim \frac{1}{\sqrt{2\pi n}} \sqrt{\frac{a+b+A+B}{(a+A)(b+B)}} \left( \frac{(a+b+A+B)^{a+b+A+B}} {(a+A)^{a+A}(b+B)^{b+B}} \right)^n . ]
Combining these three estimates gives
[ q(a,b,A,B,n) \sim \frac{C}{\sqrt{2\pi n}}E^n , ]
where
[ C= \sqrt{ \frac{(a+b)(A+B)(a+A)(b+B)} {abAB(a+b+A+B)} } ]
and
[ E= \frac{ (a+b)^{a+b}(A+B)^{A+B}(a+A)^{a+A}(b+B)^{b+B} }{ a^a b^b A^A B^B (a+b+A+B)^{a+b+A+B} }. ]
The remaining question is whether (E=1). In general it is not. Taking logarithms,
H(a,b)+H(A,B)-H(a+A,b+B), ]
where
[ H(x,y)=(x+y)\log(x+y)-x\log x-y\log y . ]
There is no identity
[ H(a,b)+H(A,B)=H(a+A,b+B) ]
for arbitrary positive (a,b,A,B). For example, with
[ a=1,\qquad b=2,\qquad A=3,\qquad B=4, ]
one obtains
[ E= \frac{3^3,7^7,4^4,6^6} {1^1,2^2,3^3,4^4,10^{10}} \approx 0.96058, ]
so (E\neq1).
Hence the actual asymptotic behaviour is
[ \boxed{ q(a,b,A,B,n) \sim \frac{C}{\sqrt{2\pi n}}E^n } ]
with (E) as above. Unless the parameters satisfy the additional condition (E=1), the quotient is not of order (n^{-1/2}); it has an exponential factor.
Therefore the exercise statement, with only the assumption
[ a,b,A,B>0, ]
is missing a hypothesis or contains a misprint. The claimed conclusion
[ q(a,b,A,B,n)=\Theta(n^{-1/2}) ]
does not hold in general. (\square)
The corrected computation shows precisely where the original proof fails: the square-root term is correct, but the exponential Stirling factor does not cancel.