I know that this series converges to $-\ln2$. I'm not looking for someone to show me that here. I've used a way of telescoping the series that gives a result of 0. I want to know what's wrong about my process, I don't care (I already know) about other ways to solve this.
$$\sum_{n=2}^\infty \ln\left(1-\frac1{n^2}\right) = \sum_{n=2}^\infty \ln\left(\frac{n^2-1}{n^2}\right) = \sum_{n=2}^\infty \left( \ln\left(n^2-1\right) - \ln\left(n^2\right)\right)$$
From that we can start telescoping:
$$= \ln3 - \ln 4 + \ln 8 - \ln 9 + \ln 15 - \ln 16 + \ln 24 - \ln 25 + \ln 35 - \ln 36 + \ln 48 - \ln49 + \ln 63 - \ln 64 + \ln 80 - \ln 81 + \,\,...$$
I separate the odd and even ones and make pairs:
$$= (\ln3 - \ln 9) + (\ln 15 - \ln 25) + (\ln 35 - \ln 49) + (\ln 63 - \ln 81) + \,\,... \\+ (-\ln 4 + \ln 8) + (-\ln 16 + \ln 24) + (-\ln36 + \ln48) + (-\ln 64 + \ln80) + \,\,... $$
$$= \ln\frac13 + \ln\frac35 + \ln\frac57 + \ln\frac79 + \,\,...\\+\ln 2 + \ln\frac32 + \ln\frac43 + \ln\frac54 + \,\,...$$
And they all cancel out from that.
$$=\ln\left(\frac{1\cdot 3\cdot 5\cdot 7\cdot\, ...}{3\cdot 5\cdot 7\cdot 9\cdot\, ...}\right) + \ln\left(\frac{2\cdot 3\cdot 4\cdot 5\cdot\, ...}{2\cdot 3\cdot 4\cdot\, ...}\right)$$
We end up with $\ln 1 + \ln 1$ which is 0. Any explanation? What did I do wrong?
Your argument should work if you carefully analyze the partial sum:
\begin{align*} \sum_{k=1}^{2n} \log\left(1-\frac{1}{k^2}\right) &= \log\left(\frac{1\cdot3\cdots(2n-1)}{3\cdot5\cdots(2n+1)}\right) + \log\left(\frac{2\cdot3\cdots(n+1)}{1\cdot2\cdots n}\right) \\ &= -\log(2n+1) + \log(n+1) \\ &= \log\left(\frac{n+1}{2n+1}\right) \\ &\xrightarrow[n\to\infty]{} \log\frac{1}{2}. \end{align*}