Telescoping with logarithms

1. $\ln\left(1+\frac1n\right)=\ln(n+1)-\ln n$, so $S_N=\ln(N+1)\to+\infty$. The series diverges, even though $\ln(1+\frac1n)\to0$: the terms tend to $0$ too slowly (they behave like $\frac1n$).

2. $1-\frac1{n^2}=\frac{(n-1)(n+1)}{n^2}$, so
$$\begin{aligned} \ln\left(1-\frac1{n^2}\right)&=\ln(n-1)-2\ln n+\ln(n+1)\\ &=\bigl[\ln(n+1)-\ln n\bigr]-\bigl[\ln n-\ln(n-1)\bigr]. \end{aligned}$$
Summing from $2$ to $N$, the brackets telescope:
$$S_N=\ln(N+1)-\ln N-\bigl(\ln2-\ln1\bigr)\longrightarrow-\ln2.$$

Remark. Exponentiating gives the infinite product $\prod_{n\ge2}\left(1-\frac1{n^2}\right)=\frac12$.