4 Summing series

4.1 The arithmetic series

Consider the finite arithmetic series with 14 terms

$\phantom{\rule{2em}{0ex}}1+3+5+\cdots +23+25+27$

A simple way of working out the value of the sum is to create a second series which is the first written in reverse order. Thus we have two series, each with the same value $A$ :

$\phantom{\rule{2em}{0ex}}A=1+3+5+\cdots +23+25+27$

and

$\phantom{\rule{2em}{0ex}}A=27+25+23+\cdots +5+3+1$

Now, adding the terms of these series in pairs

$\phantom{\rule{2em}{0ex}}2A=28+28+28+\cdots +28+28+28=28×14=392\phantom{\rule{2em}{0ex}}\text{so}\phantom{\rule{1em}{0ex}}A=196.$

We can use this approach to find the sum of $n$ terms of a general arithmetic series.

If

$\phantom{\rule{2em}{0ex}}A=\left[a\right]+\left[a+d\right]+\left[a+2d\right]+\cdots +\left[a+\left(n-2\right)d\right]+\left[a+\left(n-1\right)d\right]$

then again simply writing the terms in reverse order:

$\phantom{\rule{2em}{0ex}}A=\left[a+\left(n-1\right)d\right]+\left[a+\left(n-2\right)d\right]+\cdots +\left[a+2d\right]+\left[a+d\right]+\left[a\right]$

Adding these two identical equations together we have

$\phantom{\rule{2em}{0ex}}2A=\left[2a+\left(n-1\right)d\right]+\left[2a+\left(n-1\right)d\right]+\cdots +\left[2a+\left(n-1\right)d\right]$

That is, every one of the $n$ terms on the right-hand side has the same value: $\left[2a+\left(n-1\right)d\right].$ Hence

$\phantom{\rule{2em}{0ex}}2A=n\left[2a+\left(n-1\right)d\right]\phantom{\rule{2em}{0ex}}\text{so}\phantom{\rule{2em}{0ex}}A=\frac{1}{2}n\left[2a+\left(n-1\right)d\right].$

Key Point 2

The arithmetic series

$\left[a\right]+\left[a+d\right]+\left[a+2d\right]+\cdots +\left[a+\left(n-1\right)d\right]$
having $n$ terms has sum $A$ where:
$A=\frac{1}{2}n\left[2a+\left(n-1\right)d\right]$

As an example

$\phantom{\rule{2em}{0ex}}1+3+5+\cdots +27\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{has}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}a=1,\phantom{\rule{1em}{0ex}}d=2,\phantom{\rule{1em}{0ex}}n=14$

So   $A=1+3+\cdots +27=\frac{14}{2}\left[2+\left(13\right)2\right]=196.$

4.2 The geometric series

We can also sum a general geometric series .

Let

$\phantom{\rule{2em}{0ex}}G=a+ar+a{r}^{2}+\cdots +a{r}^{n-1}$

be a geometric series having exactly $n$ terms. To obtain the value of $G$ in a more convenient form we first multiply through by the common ratio $r$ :

$\phantom{\rule{2em}{0ex}}rG=ar+a{r}^{2}+a{r}^{3}+\cdots +a{r}^{n}$

Now, writing the two series together:

$\phantom{\rule{2em}{0ex}}G=a+ar+a{r}^{2}+\cdots +a{r}^{n-1}$

$\phantom{\rule{2em}{0ex}}rG=ar+a{r}^{2}+a{r}^{3}+\cdots a{r}^{n-1}+a{r}^{n}$

Subtracting the second expression from the first we see that all terms on the right-hand side cancel out, except for the first term of the first expression and the last term of the second expression so that

$\phantom{\rule{2em}{0ex}}G-rG=\left(1-r\right)G=a-a{r}^{n}$

Hence (assuming $r\ne 1$ )

$\phantom{\rule{2em}{0ex}}G=\frac{a\left(1-{r}^{n}\right)}{1-r}$

(Of course, if $r=1$ the geometric series simplifies to a simple arithmetic series with $d=0$ and has sum $G=na$ .)

Key Point 3

The geometric series

$a+ar+a{r}^{2}+\cdots +a{r}^{n-1}$
having $n$ terms has sum $G$ where

$\phantom{\rule{2em}{0ex}}G=\frac{a\left(1-{r}^{n}\right)}{1-r},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}r\ne 1\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}G=na,\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}r=1$

Find the sum of each of the following series:

1. $1+2+3+4+\cdots +100$
2. $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}+\frac{1}{486}$
1. In this arithmetic series state the values of $a,d,n$ :

$a=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}d=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}n=100.$

Now find the sum:

$1+2+3+\cdots +100=50\left(2+99\right)=50\left(101\right)=5050.$

2. In this geometric series state the values of $a,r,n$ :

$a=\frac{1}{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}r=\frac{1}{3},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}n=6$

Now find the sum:

$\frac{1}{2}+\frac{1}{6}+\cdots +\frac{1}{486}=\frac{1}{2}\frac{\left(1-{\left(\frac{1}{3}\right)}^{6}\right)}{1-\frac{1}{3}}=\frac{3}{4}\left(1-{\left(\frac{1}{3}\right)}^{6}\right)=0.74897$

Exercises
1. Which of the following sequences is convergent?
1. $sin\frac{\pi }{2},\phantom{\rule{1em}{0ex}}sin\frac{2\pi }{2},\phantom{\rule{1em}{0ex}}sin\frac{3\pi }{2},\phantom{\rule{1em}{0ex}}sin\frac{4\pi }{2},\phantom{\rule{1em}{0ex}}\dots$
2. $\frac{sin\frac{\pi }{2}}{\frac{\pi }{2}},\phantom{\rule{1em}{0ex}}\frac{sin\frac{2\pi }{2}}{\frac{2\pi }{2}},\phantom{\rule{1em}{0ex}}\frac{sin\frac{3\pi }{2}}{\frac{3\pi }{2}},\phantom{\rule{1em}{0ex}}\frac{sin\frac{4\pi }{2}}{\frac{4\pi }{2}},\phantom{\rule{1em}{0ex}}\dots$
2. Write the following series in summation form:
1. $\frac{ln1}{2×1}+\frac{ln3}{3×2}+\frac{ln5}{4×3}+\cdots +\frac{ln27}{15×14}\phantom{\rule{1em}{0ex}}$
2. $-\frac{1}{2×\left(1+{\left(100\right)}^{2}\right)}+\frac{1}{3×\left(1-{\left(200\right)}^{2}\right)}-\frac{1}{4×\left(1+{\left(300\right)}^{2}\right)}+\cdots +\frac{1}{9×\left(1-{\left(800\right)}^{2}\right)}\phantom{\rule{1em}{0ex}}$
3. Write out the first three terms and the last term of the following series:
1. ${\sum }_{p=1}^{17}\frac{{3}^{p-1}}{p!\left(18-p\right)}$
2. ${\sum }_{p=4}^{17}\frac{{\left(-p\right)}^{p+1}}{p\left(2+p\right)}$
4. Sum the series:
1. $-5\phantom{\rule{1em}{0ex}}-1\phantom{\rule{1em}{0ex}}+3\phantom{\rule{1em}{0ex}}+7\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+27$
2. $-5\phantom{\rule{1em}{0ex}}-9\phantom{\rule{1em}{0ex}}-13\phantom{\rule{1em}{0ex}}-17\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-37$
3. $\frac{1}{2}-\frac{1}{6}+\frac{1}{18}-\frac{1}{54}+\frac{1}{162}-\frac{1}{486}$
1. no; this sequence is $1,\phantom{\rule{1em}{0ex}}0,\phantom{\rule{1em}{0ex}}-1,\phantom{\rule{1em}{0ex}}0,\phantom{\rule{1em}{0ex}}1,\dots$ which does not converge.
2. yes; this sequence is $\frac{1}{\pi ∕2},\phantom{\rule{1em}{0ex}}0,\phantom{\rule{1em}{0ex}}-\frac{1}{3\pi ∕2},\phantom{\rule{1em}{0ex}}0,\phantom{\rule{1em}{0ex}}\frac{1}{5\pi ∕2},\dots$ which converges to zero.
1. ${\sum }_{p=1}^{14}\frac{ln\left(2p-1\right)}{\left(p+1\right)\left(p\right)}$
2. ${\sum }_{p=1}^{8}\frac{{\left(-1\right)}^{p}}{\left(p+1\right)\left(1+{\left(-1\right)}^{p+1}{p}^{2}1{0}^{4}\right)}$
1. $\frac{1}{17},\phantom{\rule{1em}{0ex}}\frac{3}{2!\left(16\right)},\phantom{\rule{1em}{0ex}}\frac{{3}^{2}}{3!\left(15\right)},\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}},\frac{{3}^{16}}{17!}$
2. $-\frac{{4}^{5}}{\left(4\right)\left(6\right)},\phantom{\rule{1em}{0ex}}\frac{{5}^{6}}{\left(5\right)\left(7\right)},\phantom{\rule{1em}{0ex}}-\frac{{6}^{7}}{\left(6\right)\left(8\right)},\phantom{\rule{1em}{0ex}}\dots \phantom{\rule{1em}{0ex}},\frac{1{7}^{18}}{\left(17\right)\left(19\right)}$
1. This is an arithmetic series with $a=-5,\phantom{\rule{1em}{0ex}}d=4,\phantom{\rule{1em}{0ex}}n=9$ . $A=99$
2. This is an arithmetic series with $a=-5,\phantom{\rule{1em}{0ex}}d=-4,\phantom{\rule{1em}{0ex}}n=9$ . $A=-189$
3. This is a geometric series with $a=\frac{1}{2},\phantom{\rule{1em}{0ex}}r=-\frac{1}{3},\phantom{\rule{1em}{0ex}}n=6$ . $G\approx 0.3745$