convergence tests for series $p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$ If the sequence: 
$p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$
Prove that the sequence 
$((n+1/2)p_n^2)^{n=\infty}_{1}$ is decreasing.
and that the series $(np_n^2)^{n=\infty}_{1}$ is convergent.
Any hints/ answers would be great. 
I'm unsure where to begin.
 A: Firstly, it should be noted that $\dfrac{p_{n+1}}{p_n}=\dfrac{2n+1}{2n+2}=\dfrac{n+\frac{1}{2}}{n+1},$ therefore, 
$$\frac{a_{n+1}}{a_n}=\frac{\left(n+1+\frac{1}{2} \right)p_{n+1}^2}{\left(n+\frac{1}{2} \right)p_{n}^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)^2}{\left(n+\frac{1}{2} \right)\left(n+1 \right)^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)}{\left(n+1 \right)^2}=\frac{n^2+2n+\frac{3}{4}}{n^2+2n+1}<1.$$
Next, $p_n$ can be rewritten as $$p_n=\frac{3}{2}\cdot \frac{5}{4}\cdot \ldots \cdot \frac{2n-1}{2n-2}\cdot \frac{1}{2n} > \frac{3}{2}\cdot \frac{1}{2n}=\frac{3}{4n}, $$
which implies
$$
p_n^2>\frac{9}{16}\cdot\frac{1}{n^2}
$$
and $$np_n^2>n\cdot\dfrac{9}{16}\cdot\dfrac{1}{n^2}=\dfrac{9}{16}\cdot\dfrac{1}{n},$$
so  $\sum\limits_{n=1}^{\infty}{np_n^2}$ diverges.
A: Hint 1: 
Show that (n+1/2)>=(n+1.5)(2n+1/2n+2)^2 for all positive integers n, then use induction to show that the first sequence is decreasing
Hint 2:
show that 1/2n<=p(n), thus 1/2n<= np(n)^2 therefore the second series diverges
