4 Summing series

4.1 The arithmetic series

Consider the finite arithmetic series with 14 terms

$1+3+5+\cdots +23+25+27$

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 $A$ :

$A=1+3+5+\cdots +23+25+27$

and

$A=27+25+23+\cdots +5+3+1$

Now, adding the terms of these series in pairs

$2A=28+28+28+\cdots +28+28+28=28\times 14=392\text{so}A=196.$

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

If

$A=\left[a\right]+\left[a+d\right]+\left[a+2d\right]+\cdots +\left[a+\left(n-2\right)d\right]+\left[a+\left(n-1\right)d\right]$

then again simply writing the terms in reverse order:

$A=\left[a+\left(n-1\right)d\right]+\left[a+\left(n-2\right)d\right]+\cdots +\left[a+2d\right]+\left[a+d\right]+\left[a\right]$

Adding these two identical equations together we have

$2A=\left[2a+\left(n-1\right)d\right]+\left[2a+\left(n-1\right)d\right]+\cdots +\left[2a+\left(n-1\right)d\right]$

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

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

As an example

$1+3+5+\cdots +27\text{has}a=1,d=2,n=14$

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

$G=a+ar+a{r}^2+\cdots +a{r}^{n-1}$

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 $r$ :

$rG=ar+a{r}^2+a{r}^3+\cdots +a{r}^n$

Now, writing the two series together:

$G=a+ar+a{r}^2+\cdots +a{r}^{n-1}$

$rG=ar+a{r}^2+a{r}^3+\cdots a{r}^{n-1}+a{r}^n$

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

$G-rG=\left(1-r\right)G=a-a{r}^n$

Hence (assuming $r\ne 1$ )

$G=\frac{a\left(1-r^n\right)}{1-r}$

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

Task!

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 $a,d,n$ :

   Answer

   $a=1,d=1,n=100.$

   Now find the sum:

   Answer

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

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

   Answer

   $a=\frac{1}{2},r=\frac{1}{3},n=6$

   Now find the sum:

   Answer

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

Exercises

1. Which of the following sequences is convergent?
   1. $sin\frac{\pi }{2},sin\frac{2\pi }{2},sin\frac{3\pi }{2},sin\frac{4\pi }{2},\dots$
   2. $\frac{sin\frac{\pi }{2}}{\frac{\pi }{2}},\frac{sin\frac{2\pi }{2}}{\frac{2\pi }{2}},\frac{sin\frac{3\pi }{2}}{\frac{3\pi }{2}},\frac{sin\frac{4\pi }{2}}{\frac{4\pi }{2}},\dots$
2. Write the following series in summation form:
   1. $\frac{ln1}{2\times 1}+\frac{ln3}{3\times 2}+\frac{ln5}{4\times 3}+\cdots +\frac{ln27}{15\times 14}$
   2. $-\frac{1}{2\times \left(1+{\left(100\right)}^2\right)}+\frac{1}{3\times \left(1-{\left(200\right)}^2\right)}-\frac{1}{4\times \left(1+{\left(300\right)}^2\right)}+\cdots +\frac{1}{9\times \left(1-{\left(800\right)}^2\right)}$
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. $-5-1+3+7\dots +27$
   2. $-5-9-13-17\dots -37$
   3. $\frac{1}{2}-\frac{1}{6}+\frac{1}{18}-\frac{1}{54}+\frac{1}{162}-\frac{1}{486}$

Answer