Proving the convergence of $\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$ How to prove that the series
$$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$
is convergent? What about finding the sum?
My attempt:
$$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$
The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum.
 A: $$log(1-\frac{1}{n^2})= \\log(\frac{n^2-1}{n^2})=\\log(\frac{n-1}{n})+log(\frac{n+1}{n})\\s_{1}+s_{2}\\s_{1}=log(\frac{2-1}{2})+log(\frac{3-1}{3})+log(\frac{4-1}{4})+...++log(\frac{n-1}{n})=\\+log(\frac{1}{2}\frac{2}{3}\frac{3}{4}...\frac{n-1}{n})\\=log(\frac{1}{n})\\$$ $$
s_{2}=log(\frac{2+1}{2})+log(\frac{3+1}{3})+log(\frac{4+1}{4})+...++log(\frac{n+1}{n})=\\log(\frac{3}{2}\frac{4}{3}\frac{5}{4}...\frac{n+1}{n})\\=log(\frac{n+1}{2})\\s_{1}+s_{2}=\\log(\frac{1}{n})+log(\frac{n+1}{2})=log(\frac{n+1}{2n}) \rightarrow log(\frac{1}{2})
 $$
A: $$\sum_{n=2}^\infty\ln (1-1/n^2)=\ln\prod_{n=2}^\infty\frac{n^2-1}{n^2}=\ln\lim_{N\to\infty}\prod_{n=2}^N\frac{n-1}n\prod_{n=2}^N\frac{n+1}n$$
Now note that
$$\frac12\cdot\frac23\cdot\ldots\frac{n+1}n=\frac1n\\
\frac32\cdot\frac43\cdot\ldots\frac n{n+1}=\frac n2$$
Therefore the answer is
$$\ln\frac12=-\ln2$$
A: $$\sum_{n=2}^{N}\log\left(1-\frac{1}{n^2}\right)=\log\prod_{n=2}^{N}\left(1-\frac{1}{N}\right)\left(1+\frac{1}{N}\right) = \log\frac{N+1}{2N}$$
so the series converges to $\color{red}{-\log 2}$.
A: Your idea is good: compute the $N$-th partial sum. You will observe that it admits a limit because it is a telescopic sum. Define indeed $a_n:= \log n$ and notice that $\sum_{j=2}^Na_{j-1}-a_j=a_1-a_N$, $\sum_{j=2}^N-a_{j}+a_{j+1}=a_{N+1}-a_2$ and $\lim_{N\to \infty}a_{N+1}-a_N=0$.
A: Let $a_n = \ln(n+1) - \ln(n)$ which means $\ln \left(1-\frac{1}{n^2} \right ) = a_n - a_{n-1}$. Its easy to see that $a_n \to 0$ as $n \to \infty$. So the given series converges and the sum is $-a_1 = -\ln(2)$
