### 3 Series

A series is the sum of the terms of a sequence. For example, the harmonic series is

$\phantom{\rule{2em}{0ex}}1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$

and the alternating harmonic series is

$\phantom{\rule{2em}{0ex}}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$

#### 3.1 The summation notation

If we consider a general sequence

$\phantom{\rule{2em}{0ex}}{a}_{1},\phantom{\rule{1em}{0ex}}{a}_{2},\phantom{\rule{1em}{0ex}}\dots ,\phantom{\rule{1em}{0ex}}{a}_{n},\phantom{\rule{1em}{0ex}}\dots$

then the sum of the first $k$ terms ${a}_{1}+{a}_{2}+{a}_{3}+\cdots +{a}_{k}$ is concisely denoted by ${\sum }_{p=1}^{k}\phantom{\rule{1em}{0ex}}{a}_{p}$ .

That is,

$\phantom{\rule{2em}{0ex}}{a}_{1}+{a}_{2}+{a}_{3}+\cdots +{a}_{k}={\sum }_{p=1}^{k}{a}_{p}$

When we encounter the expression ${\sum }_{p=1}^{k}{a}_{p}$ we let the index ‘ $p$ ’ in the term ${a}_{p}$ take, in turn, the values $1,2,\dots ,k$ and then add all these terms together. So, for example

$\phantom{\rule{2em}{0ex}}{\sum }_{p=1}^{3}{a}_{p}={a}_{1}+{a}_{2}+{a}_{3}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{\sum }_{p=2}^{7}{a}_{p}={a}_{2}+{a}_{3}+{a}_{4}+{a}_{5}+{a}_{6}+{a}_{7}$

Note that $p$ is a dummy index; any letter could be used as the index. For example ${\sum }_{i=1}^{6}{a}_{i}$ , and ${\sum }_{m=1}^{6}{a}_{m}$ each represent the same collection of terms: ${a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}+{a}_{5}+{a}_{6}$ .

In order to be able to use this ‘summation notation’ we need to obtain a suitable expression for the ‘typical term’ in the series. For example, the finite series

$\phantom{\rule{2em}{0ex}}{1}^{2}+{2}^{2}+\cdots +{k}^{2}$

may be written as ${\sum }_{p=1}^{k}{p}^{2}$ since the typical term is clearly ${p}^{2}$ in which $p=1,2,3,\dots ,k$ in turn.

In the same way

$\phantom{\rule{2em}{0ex}}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots -\frac{1}{16}={\sum }_{p=1}^{16}\frac{{\left(-1\right)}^{p+1}}{p}$

since an expression for the typical term in this alternating harmonic series is ${a}_{p}=\frac{{\left(-1\right)}^{p+1}}{p}$ .

Write in summation form the series

$\phantom{\rule{2em}{0ex}}\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots +\frac{1}{21×22}$

First find an expression for the typical term, “the ${p}^{th}$ term”:

${a}_{p}=\frac{1}{p\left(p+1\right)}$

Now write the series in summation form:

$\frac{1}{1×2}+\frac{1}{2×3}+\cdots \phantom{\rule{2em}{0ex}}+\frac{1}{21×22}={\sum }_{p=1}^{21}\frac{1}{p\left(p+1\right)}$

Write out all the terms of the series ${\sum }_{p=1}^{5}\frac{{\left(-1\right)}^{p}}{{\left(p+1\right)}^{2}}$ .

Give $p$ the values $1,2,3,4,5$ in the typical term $\frac{{\left(-1\right)}^{p}}{{\left(p+1\right)}^{2}}$ :

$-\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}-\frac{1}{{4}^{2}}+\frac{1}{{5}^{2}}-\frac{1}{{6}^{2}}$ .