# Obtaining the sum of a series

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?

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Let's start with the last one. Do you know how to combine $1-{1\over n+1}$ into a single fraction? – Gerry Myerson Nov 1 '12 at 6:37

1) As the solution states, you get that expansion by using the method of partial fractions. There are many websites that cover this topic, but here you may want to just verify the identity: \begin{align} \frac{1}{k}-\frac{1}{k+1} &= \frac{k+1}{k(k+1)}-\frac{k}{k(k+1)} \\ &= \frac{k+1-k}{k(k+1)} \\ &= \frac{1}{k(k+1)}. \end{align}

2) This is somewhat misleading. The $(1+\frac{1}{2}+\cdots+\frac{1}{n})$ does not become 1. Instead, most of the terms cancel with the $(\frac{1}{2}+\cdots+\frac{1}{n+1})$ immediately after!

3) Here's how it works: $$1-\frac{1}{n+1}=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}.$$

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Why could my brain not see that? Thanks – Gineer Nov 1 '12 at 6:55

$\frac{1}{(2)(4)}+\frac{1}{(4)(6)}+\frac{1}{(6)(8)}+...+\frac{1}{(2n)2(n+1)}$

$\implies \frac{1}{2*2}[\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+...+\frac{1}{(n)(n+1)}]$

$\implies \frac{1}{4}[\frac{2-1}{(1)(2)}+\frac{3-2}{(2)(3)}+\frac{4-3}{(3)(4)}+...+\frac{(n+1)-n}{(n)(n+1)}]$

$\implies \frac{1}{4}[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac 1 3 -\frac 1 4 + \cdots + \frac 1 n-\frac 1 {(n+1)}]$

Notice the alternate terms get cancelled out, except the first and the last terms.

$\implies \frac 1 4[1-\frac1 {n+1}]$

$\implies \frac n {4(n+1)}$

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