I am trying to evaluate
$$\sum_{n=1}^{\infty}\ln{(2n+1)n \over (n+1)(2n-1)}$$
First, we notice $${(2n+1)n \over (n+1)(2n-1)} = {2n^2+n -1 + 1 \over 2n^2+n-1} = 1 + {1 \over 2n^2+n-1}= 1 + {1 \over (n+1)(2n-1)}$$ Then, we use ${x \over 1+x} \le \ln(1+x) \le x$: $${1 \over (2n+1)n} \le \ln\left(1 + {1 \over (n+1)(2n-1)} \right) \le {1 \over (n+1)(2n-1)}$$ Now, I am going for the squeeze theorem with the partial sum $${1 \over 3} + {1 \over 10} + \dots + {1 \over (2p+1)p} \le \ln\left( 1 + {1 \over 2} \right) + \ln\left( 1 + {1 \over 9} \right) + \dots +\ln\left( 1 + {1 \over (p+1)(2p-1)} \right) \\ \le {1 \over 2} + {1 \over 9} + \dots + {1 \over (p+1)(2p-1)}$$ I get stuck on reorganizing the terms.
Hint. This may be seen a limit of two telescoping sums: $$ \sum_{n=1}^N\ln{(2n+1)n \over (n+1)(2n-1)}=\sum_{n=1}^N\left(\ln(2n+1)-\ln(2n-1)\right)-\sum_{n=1}^N\left(\ln(n+1)-\ln(n)\right). $$