Convergence of $\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$ I have to show that the series 
$\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$
converges.
I have tried Ratio Test and Cauchy Condensation Test but it didn't work for me. I tried using Comparison Test but I couldn't make an appropriate inequality for it. Could you please give me some hints. Any help will be appreciated.

Since $\log (1 + x) < x$ for non-negative $x$, we have
$$\begin {eqnarray}
0 < \sum_{n = 1}^{\infty} \log \left (\frac {(2n)^2} {(2n - 1) (2n + 1)} \right) & = & \sum_{n = 1}^{\infty} \log \left (1 + \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & < & \sum_{n = 1}^{\infty} \left ( \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & = & \frac {1} {2} \left (\sum_{n = 1}^{\infty} \frac {1} {2n - 1} - \sum_{n = 1}^{\infty} \frac {1} {2n + 1} \right) \nonumber \\ & = & \frac {1} {2} \left (\sum_{n = 0}^{\infty} \frac {1} {2n + 1} - \sum_{n = 1}^{\infty} \frac {1} {2n + 1} \right) \nonumber \\ & = & \frac {1} {2} \cdot 1 = \frac {1} {2}.
\end {eqnarray}$$
So, the series converges.
