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.
 A: I couldn't resist
We use the fact that $e^{\sum\log  (s_n)}=\prod s_n $
Setting $s_n=\frac{1}{1-\frac{1}{4n^2}}$, we find that
$$
e^{\sum_{n=1}^{\infty}\log  (s_n)}= \prod_{n=1}^{\infty} \left(1-\frac{1}{4n^2}\right)^{-1}
$$
Using the product expansion of $\sin(\pi x)$ 
$$
\frac{\pi x}{\sin(\pi x)}=\frac{1}{\prod_{n=1}^{\infty} \left(1-\frac{x^2}{n^2}\right)}
$$
and setting $x=1/2$ we immediately find that 
$$
e^{\sum_{n=1}^{\infty} \log(s_n)}=\frac{\pi}{2}
$$
and therefore 
$$
\sum_{n=1}^{\infty}\log\left(\frac{1}{1-\frac{1}{4n^2}}\right)=\log\left(\frac{\pi}{2}\right)
$$
And therefore the sum converges ;)
A: $$\frac{4 n^2}{4 n^2-1} = \frac1{1-\frac1{4 n^2}}$$
Thus,
$$\log{\left (\frac{4 n^2}{4 n^2-1} \right )} = -\log{\left ( 1-\frac1{4 n^2} \right )}$$
which, for large $n$, is approximately equal to $1/(4 n^2)$.  Thus the sum converges by the comparison test with $\zeta(2)$.
A: $0<\log (1+x)<x$ for $x>0 . $ Therefore $0<\log (\frac{4 n^2}{4 n^2-1})=\log (1+\frac{1}{4 n^2-1})< \frac{1}{4 n^2-1} < \frac{1}{2 n^2}. $ The sum $\sum (1/2 n^2)$ converges by Cauchy Condensation. Your sum therefore converges by Comparison.
(Where did that term  $\frac{1}{2 n^2}$ come from in the inequality? From the idea, with $4 n^2=x$, that $\frac{1}{x-1}<\frac{2}{x}$ if $x$ is big enough.)
A: 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.
