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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
- \(2,\ 5,\ 8,\ \dots\)
- \(107,\ 98,\ 89,\ \dots\,\).
Solution
- 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 \[ S_n = \dfrac{n}{2}\bigl(a+\ell\bigr) = \dfrac{n}{2}\bigl(2 + (3n-1)\bigr) = \dfrac{n}{2}\bigl(3n+1\bigr). \]
- 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
- \(2,\ 6,\ 18,\ \dots\)
- \(486,\ 162,\ 54,\ \dots\,\).
Solution
- Here \(a=2\) and \(r=3\), so \[ S_n = \dfrac{2(3^n-1)}{3-1} = 3^n-1. \]
- 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. |
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| Arithmetic series | \(a + (a+d) + (a+2d) + (a+3d) + \dotsb\) 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 \(a_n\). |
| Geometric sequence | \(a,\ ar,\ ar^2,\ ar^3,\ \dots\) The \(n\)th term is \(a_n = ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio. |
| Geometric series | \(a + ar + ar^2 + ar^3 + \dotsb\) 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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