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{\bf\large American Mathematical Monthly (11/99)}
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{\bf Problem 10762} [1999,864]. Let $x_1=1$ and for $m \ge 1$ let
$x_{m+1}= (m+3/2)^{-1} \sum_{k=1}^m x_k x_{m+1-k}$. Evaluate
$\lim_{m\to \infty} x_m/x_{m+1}$.
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{\it Solution by Andrew Bernoff and Karl Mahlburg (student), Harvey Mudd
College, Claremont, CA.} Define a generating function $f(z) =
\sum_{n=1}^{\infty} x_n z^n $. Multiplying both sides of the
given sum by $(m+3/2) z^{m+1}$ and summing over positive $m$
leads to a DE for $f(z)$,
$$ z f_z + \frac{1}{2}f - \frac{3}{2} z = f^2 \ .$$
Our generating function is a solution to this DE which is analytic in
a neighborhood of the origin, where $f(0)=0$.
Making the substitution $s=\sqrt{z}$, $f(z) = \frac{1}{2} +sg(s)$
yields a separable DE, $g_s = 2g^2 +3$. Selecting the
appropriate solution for $g$ yields the solution for $f$,
$$f(z) =
\frac{1}{2} \left [ 1- \sqrt{6z}\cot(\sqrt{6z}) \right ]
$$
which is meromorphic, has a removeable singularity at the origin
and poles at $ N\pi/\sqrt{6}$ for $N$ a non-zero integer.
Noting that $x_m$ is positive, we see that the required limit
is the radius of convergence of the power series for $f(z)$ at
$z=0$, given by the distance to the nearest pole in the complex plane,
that is $\lim_{m\to \infty} x_m/x_{m+1}= \pi/\sqrt{6}$.
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