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### Series

A finite series is the sum of the terms of a finite sequence. Thus, if

$a_1,\ a_2,\ \dots,\ a_n$

is a sequence of $$n$$ terms, then the corresponding series is

$a_1 + a_2 + \dots + a_n.$

The number $$a_k$$ is referred to as the $$k$$th term of the series.

We often use the sigma notation for series. For example, if we have the series

$2 + 4 + 6 + \dots + 100$

in which the $$k$$th term is given by $$2k$$, then we can write this series as

$\sum_{k=1}^{50} 2k.$

Note that the variable $$k$$ here is a dummy variable. This means that we could also write the series as

$\sum_{i=1}^{50} 2i \qquad\text{or}\qquad \sum_{j=1}^{50} 2j.$

Exercise 6

By writing out the terms, find the sum

$\sum_{k=1}^n \Bigl(\dfrac{1}{k} - \dfrac{1}{k+1}\Bigr).$

An infinite series is the formal sum' of the terms of an infinite sequence:

$a_1 + a_2 + a_3 + a_4 + \dotsb.$

For example, the sequence of odd numbers gives the infinite series $$1 + 3 + 5 + 7 + \dotsb$$.

We can sum an infinite series to a finite number of terms. The sum of the first $$n$$ terms of an infinite series is often written as

$S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^n a_k.$

This is sometimes called the $$n$$th partial sum of the infinite series.

Given a formula for the sum of the first $$n$$ terms of a series, we can recover a formula for the $$n$$th term by a simple subtraction, as follows. Starting from

\begin{align} S_n &= a_1 + a_2 + \cdots + a_{n-1} + a_n \\ S_{n-1} &= a_1 + a_2 + \cdots + a_{n-1}, \end{align}

by subtracting we obtain

$S_n - S_{n-1} = a_n.$

For example, if the sum of the first $$n$$ terms of a series is given by $$S_n= n^2$$, then the $$n$$th term is

$a_n = S_n - S_{n-1} = n^2 - (n-1)^2 = 2n-1.$

So the terms form the sequence of odd numbers. Hence, we have found a formula for the sum of the first $$n$$ odd numbers:

$1 + 3 + 5 + \dots + (2n-1) = n^2.$

In general, it can be difficult to find a simple formula for the sum of a series to $$n$$ terms. For the rest of this section, we restrict our attention to arithmetic and geometric series.

#### Arithmetic series

An arithmetic series is a series in which the terms form an arithmetic sequence. That is, each term is obtained from the preceding one by adding a constant.

The series

$1 + 2 + 3 + \dots + n$

is an arithmetic series with common difference 1. There is an easy way to find the sum of this series. We write the series forwards and then backwards:

$\begin{array}{cccccccccccc} S_n = & 1 & + & 2 & + & 3 & + & \cdots & + & (n-1) & + & n \\ S_n = & n & + & (n-1) & + & (n-2) & + & \cdots & + & 2 & + & 1. \end{array}$

Adding downwards in pairs, we obtain

$2S_n = (1+n) + (2+n-1) + (3+n-2) + \dots + (n-1+2) + (n+1).$

Each of the $$n$$ terms on the right-hand side simplifies to $$n+1$$. Thus $$2S_n = n(n+1)$$, and so we have shown that

$1 + 2 + 3 + \dots + n = \dfrac{1}{2}n(n+1).$

For example,

$1 + 2 + 3 + \dots + 100 = \dfrac{1}{2}\times 100\times 101 = 5050.$

Legend has it that the famous mathematician Gauss discovered this at the age of nine!

This trick' works for any arithmetic series, and gives a formula for the sum $$S_n$$ of the first $$n$$ terms of an arithmetic series with first term $$a_1 = a$$ and last term $$a_n = \ell$$. The formula is

$S_n = \dfrac{n}{2}\bigl(a + \ell\bigr).$

Exercise 7

Use the method of writing the arithmetic series

$a + (a+d) + (a+2d) + \dots + (\ell-d) + \ell$

forwards and backwards to derive the formula $$S_n = \dfrac{n}{2}(a + \ell)$$ given above.

Since the last term $$\ell$$ can be written as $$a_n = a + (n-1)d$$, where $$d$$ is the common difference, we also have

\begin{align*} S_n &= \dfrac{n}{2}\bigl(a+\ell\bigr) \\ &= \dfrac{n}{2}\bigl(a + a + (n-1)d\bigr) \\ &= \dfrac{n}{2}\bigl(2a + (n-1)d\bigr). \end{align*}

#### Example

Find the formula for the sum of the first $$n$$ terms of the arithmetic sequence

1. $$2,\ 5,\ 8,\ \dots$$
2. $$107,\ 98,\ 89,\ \dots\,$$.

#### Solution

1. Here $$a=2$$ and $$d=3$$, so $S_n = \dfrac{n}{2}\bigl(4 + (n-1)\times 3\bigr) = \dfrac{n}{2}\bigl(3n+1\bigr).$ Alternatively, we can find the $$n$$th term of the sequence, which is $$a_n = 3n-1$$, and use the formula $S_n = \dfrac{n}{2}\bigl(a+\ell\bigr) = \dfrac{n}{2}\bigl(2 + (3n-1)\bigr) = \dfrac{n}{2}\bigl(3n+1\bigr).$
2. Here $$a=107$$ and $$d=-9$$, so $S_n = \dfrac{n}{2}\bigl(2\times 107 + (n-1)\times -9\bigr) = \dfrac{n}{2}\bigl(223-9n\bigr).$

For both parts of the previous example, we can substitute $$n=1$$ and check this gives the first term of the series. Note that, since the formula for the sum is a quadratic, checking the three cases $$n=1$$, $$n=2$$, $$n=3$$ is sufficient to prove that the answer is correct.

Exercise 8

Sum the arithmetic series

$\log_2 3 + \log_2 9 + \log_2 27 + \dotsb$

to $$n$$ terms.

#### Geometric series

A geometric series is a series in which the terms form a geometric sequence. That is, each term is obtained from the preceding one by multiplying by a constant.

For example,

$2 + 8 + 32 + 128 + \dotsb$

is a geometric series with first term 2 and common ratio 4. The $$n$$th term is $$a_n = 2\times 4^{n-1}$$.

We can find a formula for the sum of the first $$n$$ terms of this series, again using a little trick. We multiply the series by the common ratio 4 and subtract the original, as follows. Starting from

\begin{align*} S_n &= 2 + 8 + 32 + 128 + \dots + 2\times 4^{n-1} \\ 4S_n &= 8 + 32 + 128 + \dots + 2\times 4^{n-1} + 2\times 4^{n}, \end{align*}

we subtract to obtain

$4S_n - S_n = 2\times 4^{n} - 2,$

and so

$S_n = \dfrac{1}{3}\bigl(2\times 4^n - 2\bigr) = \dfrac{2(4^n - 1)}{3}.$

This `trick' works for any geometric series, and gives a formula for the sum $$S_n$$ of the first $$n$$ terms of a geometric series with first term $$a$$ and common ratio $$r$$. The formula is

$S_n = \dfrac{a(r^n - 1)}{r - 1}, \quad\text{for } r\neq 1.$

Note that this can also be written as

$S_n = \dfrac{a(1 - r^n)}{1 - r}, \quad\text{for } r\neq 1.$

The second formula is often more convenient to use when $$r$$ lies between $$-1$$ and 1.

In the case when $$r=1$$, the sum of the series is clearly $$na$$, since all the terms are identical.

Exercise 9

Use the method of multiplying the geometric series

$a + ar + ar^2 + \dots + ar^{n-1}$

by $$r$$ and subtracting to derive the formula for $$S_n$$ given above.

#### Example

Find the formula for the sum of the first $$n$$ terms of the geometric sequence

1. $$2,\ 6,\ 18,\ \dots$$
2. $$486,\ 162,\ 54,\ \dots\,$$.

#### Solution

1. Here $$a=2$$ and $$r=3$$, so $S_n = \dfrac{2(3^n-1)}{3-1} = 3^n-1.$
2. Here $$a=486$$ and $$r=\dfrac{1}{3}$$, so $S_n = \dfrac{486\bigl(1-(\dfrac{1}{3})^n\bigr)}{1-\dfrac{1}{3}} = 729\bigl(1-(\tfrac{1}{3})^n\bigr).$

For both parts of the previous example, we can put $$n=1$$ and check that we obtain the first term of the sequence.

Exercise 10

Find the sum to $$n$$ terms of the geometric series

$\sqrt{3} + 6 + 12\sqrt{3} + \dotsb.$

#### Summary

Arithmetic series $$a,\ a+d,\ a+2d,\ a+3d,\ \dots$$ The $$n$$th term is $$a_n= a + (n-1)d$$, where $$a$$ is the first term and $$d$$ is the common difference. $$a + (a+d) + (a+2d) + (a+3d) + \dotsb$$ The sum of the first $$n$$ terms is $S_n = \dfrac{n}{2}\bigl(2a + (n-1)d\bigr),$ where $$a$$ is the first term and $$d$$ is the common difference. This can also be written $$S_n = \dfrac{n}{2}(a+\ell)$$, where $$\ell$$ is the $$n$$th term $$a_n$$. $$a,\ ar,\ ar^2,\ ar^3,\ \dots$$ The $$n$$th term is $$a_n = ar^{n-1}$$, where $$a$$ is the first term and $$r$$ is the common ratio. $$a + ar + ar^2 + ar^3 + \dotsb$$ The sum of the first $$n$$ terms is $S_n = a + ar + ar^2 + \dots + ar^{n-1} = \dfrac{a(r^n - 1)}{r - 1}, \quad\text{for } r\neq 1,$ where $$a$$ is the first term and $$r$$ is the common ratio.

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