Section 8.1 Sigma Notation
Subsection 8.1.1 Writing Complicated Sums
Sigma notation is a convenient way to write large and complicated sums. The name comes from the fact that it uses the upper case greek letter sigma: \(\Sigma\text{.}\) Sigma notation relies on an index and expresses the terms of the sum as expression of the index. Here is an example to show the general form.
\begin{equation*}
\sum_{k=1}^{34} (k^2 - 4k)
\end{equation*}
Let me define the terminology. In the example pressent, \(k\) is the index. It ranges from \(1\) to \(34\) — the numbers that appear above and below the \(\Sigma\text{.}\) These two numbers are called the bounds. The bound below the sigma tells where the index starts and the bound above the sigma tells where the index ends. The expression following the sigma is called the term. For each index from the lower bound and proceeding to the upper bound (in integer steps), I evaluate the expression. Here are some example sums to show how it works.
\begin{align*}
\sum_{k=1}^5 5k \amp = 5(1) + 5(2) + 5(3) + 5(4) + 5(5) =
5+10+15+20+25 = 75\\
\sum_{k=1}^8 \sin \left(\frac{k\pi}{2} \right) \amp = \sin
\left( \frac{(1)\pi}{2} \right) + \sin \left( \frac{(2)\pi}{2}
\right) + \sin \left( \frac{(3)\pi}{2} \right) + \sin \left(
\frac{(4)\pi}{2} \right) \\
\amp + \sin \left( \frac{(5)\pi}{2} \right)
+ \sin \left( \frac{(6)\pi}{2} \right) + \sin \left(
\frac{(7)\pi}{2} \right) + \sin \left( \frac{(8)\pi}{2}
\right)\\
\amp = 1 + 0 + -1 + 0 + 1 + 0 + -1 + 0 = 0\\
\sum_{k=1}^{36} k \amp = 1 + 2 + 3 + 4 + \ldots + 35 + 36 =
666
\end{align*}
Subsection 8.1.2 Manipulating Sums
There are some important rules for manipulating sigma notation. First, sums are (unsurprisingly) linear, so I can split them up over addition/subtraction and pull out constants.
\begin{align*}
\sum_{k=1}^n 6k \amp = 6 \sum_{k=1}^n k\\
\sum_{k=1}^n (3k^2 + 9k) \amp = 3 \sum_{k=1}^n k^2 + 9
\sum_{k=1}^n k
\end{align*}
I can combine sums if the indices match.
\begin{equation*}
\sum_{k=1}^{15} k^2 + \sum_{k=1}^{15} 4k = \sum_{k=1}^{15}
(k^2 + 4k)
\end{equation*}
If I want to combine sums when the indices don’t match, I have to adjust the sums. There are two main methods of adjusting the sum. First, I can take out terms.
\begin{equation*}
\sum_{k=1}^{15} k^2 + \sum_{k=3}^{15} 4k = 1^2 + 2^2 +
\sum_{k=3}^{15} k^2 + \sum_{k=3}^{15} 4k = 5 + \sum_{k=3}^{15}
(k^2 + 4k)
\end{equation*}
I can also shift the indices. This might seem tricky, but it’s easy to remember if you think of balance: whatever I do to the index, I have to do the opposite to the expression. Here I shift the index by \(+2\text{,}\) so the expression is shifted by \(-2\text{.}\)
\begin{equation*}
\sum_{k=3}^{17} 4k = \sum_{k=1}^{15} 4(k+2)
\end{equation*}
I can use these two manipulations to combine sums if the indices don’t initially line up.
\begin{equation*}
\sum_{k=1}^{15} k^2 + \sum_{k=3}^{17} 4k = \sum_{k=1}^{15}
k^2 + \sum_{k=1}^{15} 4(k+2) = \sum_{k=1}^{15} (k^2 + 4k +
8)
\end{equation*}
Subsection 8.1.3 Some Important Sums
Finally, here are the values for some common sums. These expressions will be used, along with the previously stated rules, to calculate integrals in the next section.
\begin{align*}
\sum_{k=1}^n 1 \amp = n\\
\sum_{k=1}^n k \amp = \frac{n(n+1)}{2}\\
\sum_{k=1}^n k^2 \amp = \frac{n(n+1)(2n+1)}{6}\\
\sum_{k=1}^n k^3 \amp = \left(\frac{n(n+1)}{2} \right)^2
\end{align*}
The last sum establishes a interesting result: the sum of the first \(n\) cubes is always a square. You are welcome to check for yourself: the first few sums are \(1\text{,}\) \(9\text{,}\) \(36\text{,}\) \(100\) and \(225\text{,}\) all of which are square.
