TITLE: convergence of series with $k!$
QUESTION [7 upvotes]: check if the following series converges:
$\sum\limits_{k=1}^{\infty} (-1)^k \dfrac{(k-1)!!}{k!!}$
where $k!!=k(k-2)(k-4)(k-6)...$
I came across this exercise while going trough some old exams. I'm pretty sure we have to bound the sequence and apply Leibniz-criteria but after a while i gave up. If you have a little Hint for me just to get me strted , that would be great.

REPLY [3 votes]: Set $a_k = (k-1)!!/k!!$, so that we're looking at the series $\sum_{k=1}^\infty (-1)^k a_k$. Notice that the odd terms are exactly the normalized central binomial coefficients:
$$
a_{2n} = \frac1{4^n} \binom{2n}n.
$$
By known asymptotics, $a_{2n} \sim 1/\sqrt{\pi n}$ as $n$ grows large. Similarly,
$$
a_{2n-1} = \frac1{2na_{2n}} \sim \frac{\sqrt{\pi n}}{2n} = \frac{\sqrt\pi}{2\sqrt n}
$$
as $n$ grows large. Therefore
\begin{align*}
\sum_{k=1}^{2N} (-1)^k a_k &= \sum_{n=1}^N ( -a_{2n-1}+a_{2n} ) \\
&\approx  \sum_{n=1}^N \bigg( {-}\frac{\sqrt\pi}{2\sqrt n}+\frac1{\sqrt{\pi n}} \bigg) \sim c\sqrt N
\end{align*}
where $c = -\sqrt\pi+2/\sqrt\pi < 0$. So the sum diverges to $-\infty$.