Sums of telescoping series

The general term is decomposed into partial fractions: looking for $\frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$ one finds $A=1$ and $B=-1$. The series is therefore telescoping with $b_n=\frac{1}{n}$: in the partial sum each term $-\frac{1}{n+1}$ cancels with the term $+\frac{1}{n+1}$ of the next summand, and the first and the last remain. Since $\frac{1}{N+1}\to 0$, the partial sums tend to $1$, therefore the series converges with sum $1$.

The partial fraction decomposition gives $A=\frac{1}{2}$ and $B=-\frac{1}{2}$, therefore $\frac{1}{2}$ is factored out. Here the cancellation happens at distance two: the term $-\frac{1}{n+2}$ cancels with the $+\frac{1}{n+2}$ that appears two summands later. Hence the first two positive terms, $1$ and $\frac{1}{2}$, survive, together with the last two negative ones, $-\frac{1}{N+1}$ and $-\frac{1}{N+2}$. The latter tend to $0$, therefore the partial sums tend to $\frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4}$.

For $n\ge 2$ the argument of the logarithm is positive, therefore the general term is defined. One factors $1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{(n-1)(n+1)}{n^2}$ and splits the logarithm of the product into the sum of the logarithms. Setting $b_n=\ln \frac{n-1}{n}$ one gets $b_{n+1}=\ln \frac{n}{n+1}$, therefore the general term is exactly $b_n-b_{n+1}$ and the series is telescoping. The partial sum equals $b_2-b_{N+1}$; since $\frac{N}{N+1}\to 1$ and the logarithm is continuous, $b_{N+1}\to \ln 1=0$, hence the series converges with sum $\ln \frac{1}{2}=-\ln 2$.