\section{Proof that \boldmath{$c_v$} tends to 1}
The following Lemma \ref{limit:c} outlines the simple steps in our proof:
\begin{lemma} \label{limit:c}Let
$c_v=\frac{\sqrt{2} \Gamma(\frac{v+1}{2})}
{\sqrt{v}\Gamma(\frac{v}{2})}$, $v=1,2,\cdots$.
Then
\begin{description}
\item[i)] $\;\;c_vc_{v+1}=\sqrt{\frac{v}{v+1}}$ increases to 1 as $v$ tends
to $\infty$.
\item[ii)] $\;c_{v+1}/c_{v-1}=\frac{v}{\sqrt{v^2-1}} > 1$
for $v \geq 2$.
\item[iii)] $c_v \leq 1$ for all $v \geq 1$.
\item[iv)] Both sequences $\{c_{2v}\}$ and $\{c_{2v+1}\}$ increase to
1.
\item[v)] $\;\;\lim_{v\rightarrow \infty}c_v=1$.
\end{description}
\end{lemma}
\begin{proof}
Both i) and ii) follow easily from the identity
$\Gamma(x+1)=x\Gamma(x)$. For a sample of size n from a normal
distribution with variance $\sigma^2$, we noted that $E(S)=c_v\sigma$,
where $v=n-1$. But $0\leq Var(S)=E(S^2)-\{E(S)\}^2=
\sigma^2-\{E(S)\}^2$. This implies that $0\leq E(S) \leq \sigma$
and so iii) is true (one could show $0< E(S) < \sigma$ when sampling
from a nondegenerate distribution with finite variance so $c_v < 1$
each v ). Next observe that ii) and iii) imply that both $\{c_{2v}\}$
and $\{c_{2v+1}\}$ are bounded, increasing sequences, and so have
finite limits, say $L_E$ and $L_O$, respectively, with both $L_E$ and
$L_O$ in the interval $(0,1]$. But i) indicates that $\lim_{v
\rightarrow \infty}c_{2v}c_{2v+1}=L_EL_O=1$, and so $L_E=L_O=1$ and
iv) is true. Finally, if both sequences $\{c_{2v}\}$ and
$\{c_{2v+1}\}$ converge to 1, it is easy to see that $\{c_{v}\}$ also
converges to 1.
\end{proof}