To what number the telescoping series converges,
$$\sum_{n=1}^\infty \ln\left( \frac{(n+1)^2}{n(n+2)} \right)$$
I already applied the log property of $$\sum_{n=1}^\infty \ln((n+1)^2) - \ln(n)(n+2)= \sum_{n=1}^\infty 2\ln(n+1) - \ln(n) - \ln(n+2)$$
we know that $$2\ln(n+1) = \ln(n+1) + \ln(n=1)$$
re-writing as two telescoping series:
$$-\sum_{n=1}^\infty \ln(n) - \ln(n+1) + \sum_{n=1}^\infty \ln(n+1) - \ln(n+2)$$
The limit of the partial sum will be:
$$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \ln(2) - \ln(1) - \ln(n+2) + \ln(n+1) \\ = \lim_{n \to \infty}\left( ln(2) + \ln\left(\frac{n+1}{n+2}\right) \right) \\ = \ln(2) $$
i'd like to know if i'm correct bc the awnser i got is the same as the book
PS: sorry for my poor formatting and English