Example 9.3.
\(\displaystyle\sum_{n=1}^{5} 2^{-n}\) represents the sum
\begin{equation*}
\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +
\frac{1}{16} + \frac{1}{32}.
\end{equation*}
Unit 9.2 Finite series
Let’s talk for a minute about a notation you have likely seen before. It is called the "Sigma" notation because that’s the name of the Greek letter \(\Sigma\text{.}\) The notation involves an indexing variable which runs between a lower limit and an upper limit. The lower and upper limits are required to be integers
. If the indexing variable is \(n\text{,}\) the lower limit is \(L\text{,}\)the upper limit is \(U\) and the general term is \(b_n\text{,}\)the summation looks like \(\displaystyle\sum_{n=L}^U b_n\text{.}\) What this means is to add together all the values of \(b_n\) starting with \(n=L\) and ending with \(n=U\text{.}\)
Aside
The summand, as you can see, is usually a function of the indexing variable; otherwise, the summand would not change from term to term. There may be other variables, for example \(\displaystyle\sum_{k=3}^6 k x\) evaluates to \(3x + 4x + 5x + 6x\text{,}\) which is equal to \(18x\text{.}\) Note that this other variable \(x\) persists when the sum is evaluated. It is a free variable. On the other hand, the index of summation, \(k\) in this case, is a bound variable. It runs over a set of values (in this case 3 to 6) and does not appear in the final value.
Checkpoint 138.
In the sum \(\displaystyle\sum_{k=1}^n \frac{C}{k}\text{,}\)which of the variables \(k\text{,}\) \(n\) and \(C\) are free and which are bound?
When a sequence is summed, the result is called a series (the plural is also series).
Example 9.4.
The sum \(\displaystyle\sum_{n=5}^{19} \frac{3}{n-2}\) represents a series with 15 terms because there are 15 integers in the range from 5 to 19. Informally, we might write this sum by writing the first few terms and the last term, with dots in between (traditionally the dots are centered for series, as opposed to at the bottom of the line for sequences). Thus we would write \({\frac{3}{3} + \frac{3}{4} + \cdots
+ \frac{3}{17}}\text{,}\) assuming this conveyed enough information for the reader to understand the precise sum.
Of course there is no reason why the index should go from 5 to 19. There have to be fifteen terms, but why not write the sum with the index going from 1 to 15? Then it would look like
\begin{equation*}
\displaystyle\sum_{n=1}^{15} \frac{3}{n+2} \, .
\end{equation*}
Another natural choice is to let the index run from 0 to 14:
\begin{equation*}
\displaystyle\sum_{n=0}^{14} \frac{3}{n+3} \, .
\end{equation*}
All three of these formulas represent the exact same sum.
