Authors: Douglas Wilhelm Harder and Prof. Hiren Patel
Very often in engineering and mathematics, it is necessary to calculate a sum of values:
$0 + 1 + 2 + 3 + \cdots + 100$
$1.3 + 1.4 + 1.5 + 1.6 + \cdots + 3.5 + 3.6$
$1 + 0.5 + 0.25 + 0.125 + 0.0625 + \cdots + 0.0009765625$
$\cos(0) + \cos(1) + \cos(2) + \cdots + \cos(100)$
The sum of a sequence of values is said to be a series, so the three above sums are also described as series. In general, we will write a series as
$a_0 + a_1 + a_2 + \cdots + a_n$
where $a_0$ is the initial term in the series, and $a_n$ is the last term.
There is a serious weakness to the above visual descriptions of series, as it is up to the reader to guess what the intermediate values are. For example, while the first series appears to be the sum of the positive integers up to $100$, that is just a reasonable guess. What happens if the author actually meant
$0 + 1 + 2 + 3 + 4 + 5 + 6 + 98 + 99 + 100$?
One common joke amoung mathematicians is: "What is the missing number in the sequence $0, 1, ?, 3, 4$?" The correct answer is, of course, $42$, as the numbers are obviously the roots of the polynomial $x^5 - 50 x^4 + 355 x^3 - 810 x^2 + 504 x$.
Now, most of us would cry foul, but it's more fundamentally a weakness of the representation.
A more compact and clearer way of representing a series is to use summation notation, where we write
$a_0 + a_1 + a_2 + \cdots + a_n = \sum_{k = 0}^n a_k$
where the index $k$ takes on the integer values from, in this case, $0$ all the way up to $n$, and these are then summed togeter. The above four series can thus be written as
$0 + 1 + 2 + 3 + \cdots + 100 = \sum_{k = 0}^{100} k$
$1.3 + 1.4 + 1.5 + 1.6 + \cdots + 3.5 + 3.6 = \sum_{k = 0}^{23} (1.3 + 0.1k)$
$1 + 0.5 + 0.25 + 0.125 + 0.0625 + \cdots + 0.0009765625 = \sum_{k = 0}^{10} 0.5^k$
$\cos(0) + \cos(1) + \cos(2) + \cdots + \cos(100) = \sum_{k = 0}^{100} \cos(k)$
There is no reason that the index starts at $0$, so you could have:
$21 + 22 + 23 + 24 + \cdots + 100 = \sum_{k = 21}^{100} k$
$1.3 + 1.4 + 1.5 + 1.6 + \cdots + 3.5 + 3.6 = \sum_{k = 13}^{23} 0.1k$
$0.25 + 0.125 + 0.0625 + \cdots + 0.0009765625 = \sum_{k = 2}^{10} 0.5^k$
$\cos(83) + \cos(84) + \cos(85) + \cdots + \cos(100) = \sum_{k = 83}^{100} \cos(k)$
Write the following series using summation notation where you should make an intelligent guess as to what the terms in the series are:
$7 + 9 + 11 + 13 + \cdots + 101$
$-1.3 + 1.4 - 1.5 + 1.6 - 1.7 + \cdots - 3.5 + 3.6$
$1 + 10 + 100 + 10000 + \cdots + 1000000000000$
$3^2 + 4^2 + 5^2 + 6^2 + \cdots + 15^2$
$f(9) + f(8) + f(7) + \cdots + f(-3)$
There are a few properties of the summation notation that you should be familiar with already.
For example, you know that:
$(a_0 + b_0) + (a_1 + b_1) + (a_2 + b_2) + (a_3 + b_3) + (a_4 + b_4) + (a_5 + b_5) + (a_6 + b_6)$
is equal to
$(a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6) + (b_0 + b_1 + b_2 + b_3 + b_4 + b_5 + b_6)$.
For example, $(3 + 12) + (4 + 14) + (5 + 16) + (6 + 18) = (3+4+5+6) + (12+14+16+18)$. The same holds true for summation notation:
$\sum_{k = m}^n(a_k + b_k) = \left(\sum_{k = m}^n a_k\right) + \left( \sum_{k = m}^n b_k\right)$.
Similarly, you know that
$ab_0 + ab_1 + ab_2 + ab_3 + ab_4 + ab_5 + ab_6 = a(b_0 + b_1 + b_2 + b_3 + b_4 + b_5 + b_6)$;
for example, $2\cdot 5 + 2\cdot 6 + 2\cdot 7 + 2\cdot 8 = 2(5+6+7+8)$. This is also true for summation notation:
$\sum_{k = m}^n ab_k = a\sum_{k = m}^n b_k$.
You are already aware that if we have the series
$a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} + a_{11}$
that this equals the sum of the two series
$(a_5 + a_6 + a_7 + a_8) + (a_9 + a_{10} + a_{11})$
and therefore
$\sum_{k = \ell}^n a_k = \sum_{k = \ell}^{m - 1} a_k + \sum_{k = m}^n a_k$.
This can be used, for example, having
$\sum_{k = 0}^n a_k = \sum_{k = 0}^{m - 1} a_k + \sum_{k = m}^n a_k$.
then
$\sum_{k = m}^n a_k = \sum_{k = 0}^n a_k - \sum_{k = 0}^{m - 1} a_k$.
We will describe counting, arithmetic and geometric series.
Finally, there are some very simple sums: the series
$\sum_{k = m}^n 1$
sums the value $1$ a total of $n - m + 1$ times, so this must be equal to $n - m + 1$:
$\sum_{k = m}^n 1 = n - m + 1$.
The $+ 1$ is important, because if you are summing $1 + 1 + 1 + 1 + 1$ five times, this could be
$\sum_{k = 0}^4 1$ or $\sum_{k = 1}^5 1$ or $\sum_{k = 98}^{102} 1$.
and in each case $4 - 0 + 1 = 5 - 1 + 1 = 102 - 98 + 1 = 5$.
Another important sum is the arithmetic series
$\sum_{k = 0}^n k = \frac{n(n + 1)}{2}$.
This formula will appear in many of your classes. This says, for example, that $0 + 1 + 2 + 3 + 4 + 5 + 6$ is equal to $\frac{6 \cdot 7}{2} = 21$, and that $0 + 1 + 2 + 3 + 4 + \cdots + 100$ is equal to $\frac{100 \cdot 101}{2} = 5050$.
Another important series is the geometric series
$\sum_{k = 0}^n r^k = \frac{1 - r^{n + 1}}{1 - r}$.
This formula will also appear in many of your classes. This says, for example, that $1 + 2 + 4 + 8 + 16 + 32 + 64$ is equal to $\frac{1 - 2^7}{1 - 2} = \frac{-127}{-1} = 127$, and that $1 + 0.5 + 0.25 + 0.125 + 0.0625 + \cdots + 0.5^{100}$ is equal to $\frac{1 - 0.5^{101}}{1 - 0.5} = \frac{0.99999999999999999999999999999960556954738949409729413571735860688516339678244548849761486053466796875}{0.5}$ which is a value very close to but just less than $2$.