<s>[INST] What is the order of magnitude of $\sum\limits_{n=1}^kn\binom{k}{n}\frac{(2k-2n-1)!!}{(2k-1)!!}$ as $k\to\infty$? What is the order of magnitude of the function $f(k)$ below in the limit as $k \rightarrow \infty$? Does it diverge, converge to a positive limit, or converge to zero?
$$
f(k) = \sum\limits_{n=1}^{k} n \binom{k}{n} \frac{(2k - 2n - 1)!! }{(2k-1)!! }
$$
Here $!!$ represents the double factorial. Does anyone have an hint?
 [/INST] We can write your $f(k)$ in the form
$$
\sum\limits_{n = 1}^k {\frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}} .
$$
Suppose that $k$ is large and fixed and consider
$$\tag{$*$}
\frac{1}{{(n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}
$$
The logarithmic derivative for $2\leq n\leq k$ satisfies
$$
\psi (k - n + 1) - \psi \!\left( {k - n + \tfrac{1}{2}} \right) - \psi (n) < \frac{2}{{k - n + 1}} - \log \left( {n - \tfrac{1}{2}} \right)<0,
$$
i.e., the sequence starts to decrease after $n=2$. For $n=1,2$ the terms of $(*)$ are
$$
\frac{{2k}}{{2k - 1}},\quad \frac{{2k(2k - 2)}}{{(2k - 1)(2k - 3)}}
$$
which are at most $2$, say, for large $k$. Thus,
$$
0 \leq \frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma\! \left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}} \le 2 \times \frac{1}{{2^n }}
$$
for all large $k$ and any $n\geq 1$. Consequently, we can apply Tannery's theorem and Stirling's formula, to deduce
\begin{align*}
& \mathop {\lim }\limits_{k \to  + \infty } \sum\limits_{n = 1}^k {\frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma\! \left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}} 
\\ &
 = \sum\limits_{n = 1}^\infty  {\frac{1}{{2^n (n - 1)!}}\mathop {\lim }\limits_{k \to  + \infty } \frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}}  = \sum\limits_{n = 1}^\infty  {\frac{1}{{2^n (n - 1)!}}}  = \frac{{\sqrt e }}{2}.
\end{align*}
</s>