Given the following example:
Obtain the Sum of the series $\frac{1}{(2)(4)}+\frac{1}{(4)(6)}+\frac{1}{(6)(8)}+...+\frac{1}{(2n)2(n+1)}=\sum_{k=1}^n{\frac{1}{4k(k+1)}}=\frac{1}{4}\sum_{k=1}^n{\frac{1}{k(k+1)}}$ Solution: Resolve $\frac{1}{k(k+1)}$ into partial fractions to get $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ Then
$\begin{align} S_n&=\frac{1}{4}\left[\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k+1}\right]\\ &= \frac{1}{4}\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n+1}\right)\right]\\ &=\frac{1}{4}\left[1-\frac{1}{n+1}\right]\\ &=\frac{1}{4}\frac{n}{n+1} \end{align}$
My questions then are:
1. How do we get $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$. Where does the minus come from?
2. How can we simplifi $\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}\right)$ to be equal to $1$? I get that if n tends to infinity that $\frac{1}{n}$ would become effectively 0, but what about the initial big fractions? They are very much bigger than 1 if you add even just a few of them together.
3. How do we go from the second last line to the last line? What happens to the minus sign? Where does the n in the numerator come from?