calculus - is $sumlimits _{n=1}^{infty}lnleft(frac{left(n+1right)^{2}}{nleft(n+2right)}right)

I don't know why but I'm having a hard time determining whether this series
$$
\sum\limits _{n=1}^{\infty}\ln\left(\frac{\left(n+1\right)^{2}}{n\left(n+2\right)}\right)
$$
converges to a real limit.

I did try to break it down according to $\ln$ identities.
$$
\ln\left(\frac{\left(n+1\right)^{2}}{n\left(n+2\right)}\right)=\ln\left(\left(n+1\right)^{2}\right)-\ln\left(n\left(n+2\right)\right)=2\ln\left(n+1\right)-\ln n-\ln\left(n+2\right)
$$
and then tried to increase it to get to a series that converges to a real number:
$$
2\ln\left(n+1\right)-\ln n-\ln\left(n+2\right)\leq 2\ln\left(n+1\right)-\ln n-\ln\left(n+1\right)=\ln\left(n+1\right)-\ln n
$$
so $S_n$ will be like that
$$

S_n=\ln2-\ln1+\ln3-\ln2+\dots+\ln(n+1)-\ln n \\=\ln (n+1)-\ln1=\ln (n+1)
$$
and so $\lim_{n\to\infty}S_n=\infty$

I know this series does converge to a real number (Well according to Wolfram Alpha :) $\ $)

Any help would be appreciated.

Answer

Hint:

Theorem: Let $\lim_{n\to\infty}n^pu_n=A$. $$\sum u_n$$ converges if $p>1$ and $A$ is finite.

Now take $p=2$.