Convergence/divergence of: $\sum_{n = 2}^\infty \ln\left( 1-\frac{1}{n^2}\right)$? How to prove the convergence/divergence of:
$$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right)$$ ?

I've tried:
$$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right) = \sum_{n = 2}^{+\infty} \ln\left(\frac{n^2-1}{n^2}\right) $$
$$= \sum_{n = 2}^{+\infty} [\ln(n^2-1) + \ln(n^2)] $$
$$= [\ln(3)-\ln(4)]+[\ln(8)-\ln(9)]+[\ln(15)-\ln(16)]+\cdots$$
But I can't see a pattern.

Thought about finding an upper limit as well, but since $$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right)$$ isn't monotonic, then (monotonic + bounded => converges) or Squeeze theorem doesn't help.
 A: We have a telescopic product:
$$\prod_{n=2}^{N}\left(1-\frac{1}{n^2}\right) = \prod_{n=2}^{N}\frac{n-1}{n}\prod_{n=2}^{N}\frac{n+1}{n} = \frac{1}{N}\cdot\frac{N+1}{2}\tag{1}$$
hence by taking logarithms and sending $N\to +\infty$ it follows that:
$$ \sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right) = \color{red}{-\log 2}.\tag{2}$$
A: You should simplify further
$$ \sum_{n = 2}^{\infty} \left(\ln(n^2-1) + \ln(n^2)\right)$$
$$=\sum_{n = 2}^{\infty} \left(\ln((n-1)(n+1)+\ln(n^2)\right)$$
$$=\sum_{n = 2}^{\infty}\left(\ln(n-1)+\ln(n+1)+2\ln(n)\right)$$
so we get
 $$\sum_{n = 2}^N \ln\left( 1-\frac{1}{n^2}\right)=-\ln(2)-\ln(N)+\ln(N+1)=-\ln 2+\ln(1+\frac{1}{N})$$
This is similar to the product solution of Jack D'Aurizio
A: Here is a more general Comparison of  $\sum_{n = 2}^∞ a_k$  and $ \sum_{n = 2}^∞ ln( 1- a_k)$ 
Suppose that for all $ k \in \mathbb N$, $0<a_k<1$.Try to prove the following exercise:
(a) $ a_k \to 0$ $ \iff $ $ln(1-a_k) \to 0$
(b) $\sum_{n = 2}^∞ a_k$ converges $ \iff $ $ \sum_{n = 2}^∞ ln( 1- a_k)$ converges.[Use the limit comparison test,keeping L'Hopital's rule in mind].Please let me know if you need some help for solving this exercise.
A: \begin{array} \\
\displaystyle\sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) &=& \displaystyle\sum_{n=2}^N \ln\left(\frac{n^2-1}{n^2}\right) \\
&=& \displaystyle\sum_{n=2}^N \ln\left(n^2-1\right)-\ln\left(n^2\right) \\
&=& \displaystyle\sum_{n=2}^N \ln\left((n-1)(n+1)\right)-2\ln n \\
&=& \displaystyle\sum_{n=2}^N \ln(n-1) + \ln(n+1) - 2\ln n \\
&=& \displaystyle [\ln(1)+\ln(3)-2\ln(2)] + [\ln(2)+\ln(4)-2\ln(3)] + [\ln(3)+\ln(5)-2\ln(4)] + \cdots \\
&&+ [\ln(N-2) + \ln(N) - 2 \ln(N-1)] + [\ln(N-1) + \ln(N+1) - 2 \ln(N)]
\end{array}
You can check that most of the terms cancel (or prove it by induction), and you have:
$$\displaystyle\sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) = -\ln(2) + \ln(N+1) - \ln(N) = -\ln(2) + \ln\left(1 + \frac{1}{N}\right)$$
And thus:
$$\displaystyle\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right) = \lim_{N\rightarrow\infty} \sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) = \lim_{N\rightarrow\infty} \left(-\ln(2) + \ln\left(1 + \frac{1}{N}\right)\right) = -\ln(2)$$
A: Hint: You can show that $\lim_{x \to 0} -x / \log (1-x) = 1$ and then use anything you might know about $p$-series.
