<s>[INST] Convergence and evaluation of $\sum_{n=1}^{\infty}\ln {(2n+1)n \over (n+1)(2n-1)}$ 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.
 [/INST] Hint. This may be seen a limit of two telescoping sums:
$$
\sum_{n=1}^N\ln{(2n+1)n \over (n+1)(2n-1)}=\sum_{n=1}^N\left(\ln(2n+1)-\ln(2n-1)\right)-\sum_{n=1}^N\left(\ln(n+1)-\ln(n)\right).
$$
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