<s>[INST] 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.
 [/INST] \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:
$$\displaystyle\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right) = \lim_{N\rightarrow\infty} \sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) = \lim_{N\rightarrow\infty} \left(-\ln(2) + \ln\left(1 + \frac{1}{N}\right)\right) = -\ln(2)$$
</s>