Introduction to Series Lesson

In the last lesson, we learned that a sequence is a function that computes an ordered list. A series is just the sum of the terms of a sequence. Let's suppose Jamie has a job at the local bakery and her starting salary is $50,000. After each year of employment, Jamie will get a 10% raise. $$a_{n + 1} = 1.1 \cdot a_{n}$$ We could use the above to obtain the terms of the sequence that describes Jamie's salary over a five-year period. $$a_{1} = 50{,}000$$ $$a_{2} = 1.1 \cdot 50{,}000 = 55{,}000$$ $$a_{3} = 1.1 \cdot 55{,}000 = 60{,}500$$ $$a_{4} = 1.1 \cdot 60{,}500 = 66{,}550$$ $$a_{5} = 1.1 \cdot 66{,}550 = 73{,}205$$ When a sequence has the terms a₁, a₂, a₃,..., then we can use S_n to represent the sum of the first n terms. In our example above, we have Jamie's salary for each year for a five-year period. We could use S₅ to represent the sum of the first five terms. $$S_{5} = 50{,}000 + 55{,}000 + 60{,}500 + 66{,}550 + 73{,}205 = 305{,}255$$ Let's look at an example.
Example #1: Find the sum of the first three terms of the given sequence. $$a_n=42 - 7n$$ First, let's find a₁, a₂, and a₃: $$a_1=42 - 7(1)=42 - 7 = 35$$ $$a_2=42 - 7(2)=42 - 14 = 28$$ $$a_3=42 - 7(3)=42 - 21 = 21$$ Now, let's find our sum. We will use S₃ to denote the sum of the first three terms of our sequence: $$S_3=35 + 28 + 21=84$$

Summation Notation

When we work with series, we generally use sigma notation, which is also known as summation notation. The Greek capital letter sigma "Σ" is used to indicate a sum.

A Finite Series is an expression of the form:

$$S_{n} = a_1 + a_2 + \cdots +a_n = \sum_{i=1}^{n}a_{i}$$

An Infinite Series is an expression of the form:

$$S_{\infty} = a_1 + a_2 + \cdots + a_n + \cdots = \sum_{i=1}^{\infty}a_{i}$$ The "i" placed at the bottom is not the imaginary unit. This is known as the index of summation. The index of summation or just the index for short is used as a dummy variable or simply a counter. You might also see letters such as "j" or "k" used for the same purpose. The index of summation is set equal to the lower limit of summation. This lower limit of summation gives us the starting index value or the value of the index that is used to obtain the first term. The "n" placed at the top is known as the upper limit of summation or the ending index value. With an infinite series, we have "∞" instead of "n" since there is no ending index value.
If i = 1, then you can read the expression as: the sum as "i" goes from 1 to "n". In other words, you would start with i = 1, and then i = 2, i = 3, and so on. This process would continue until we get to i = n. When we sum all of the terms together, we get the sum of the series or simply the series total. It's worth noting that the lower limit of summation can be an integer other than 1.

Summation Properties

If a₁, a₂,..., a_n and b₁, b₂,..., b_n are two sequences, and c is a constant, then for every positive integer n: $$1) \, \sum_{i=1}^{n}c=cn$$ $$2) \, \sum_{i=1}^{n}ca_i=c \sum_{i=1}^{n}a_i$$ $$3) \, \sum_{i=1}^{n}(a_i + b_i) =\sum_{i=1}^{n}a_i + \sum_{i=1}^{n}b_i$$ $$4) \, \sum_{i=1}^{n}(a_i - b_i) =\sum_{i=1}^{n}a_i - \sum_{i=1}^{n}b_i$$

Summation Rules

The following rules will greatly speed up our work in this section. $$1) \, \sum_{i=1}^{n}i=\frac{n(n + 1)}{2}$$ $$2) \, \sum_{i=1}^{n}i^2=\frac{n(n + 1)(2n + 1)}{6}$$ $$3) \, \sum_{i=1}^{n}i^3=\frac{n^2(n + 1)^2}{4}$$ Let's look at some examples.
Example #2: Evaluate each series. $$\sum_{i = 1}^{13} 4i$$ Here, we can use property #2: $$\sum_{i=1}^{n}ca_i=c \sum_{i=1}^{n}a_i$$ In this case, c = 4 and a_i = i. $$\sum_{i = 1}^{13} 4i = 4\sum_{i = 1}^{13} i$$ To finish things up, we can use rule #1: $$\sum_{i=1}^{n}i=\frac{n(n + 1)}{2}$$ Since our n-value is 13, we obtain the following: $$4\sum_{i = 1}^{13} i = 4 \cdot \frac{13(13 + 1)}{2} = 364$$ Example #3: Evaluate each series. $$\sum_{i=1}^{7}(5i^2 + 2)$$ Here, we can use property #3: $$\sum_{i=1}^{n}(a_i + b_i) =\sum_{i=1}^{n}a_i + \sum_{i=1}^{n}b_i$$ In this case, a_i = 5i² and b_i = 2. $$\sum_{i=1}^{7}(5i^2 + 2) = \sum_{i=1}^{7}5i^2 + \sum_{i=1}^{7}2$$ Let's also use property #2: $$\sum_{i=1}^{n}ca_i=c \sum_{i=1}^{n}a_i$$ In this case, c = 5 and a_i = i² $$\sum_{i=1}^{7}5i^2 + \sum_{i=1}^{7}2 = 5\sum_{i=1}^{7}i^2 + \sum_{i=1}^{7}2$$ Let's take these on one at a time. We will begin by using rule #2: $$\sum_{i=1}^{n}i^2=\frac{n(n + 1)(2n + 1)}{6}$$ In this case, our n-value is 7. Let's plug in and simplify. $$5\sum_{i=1}^{7}i^2=5 \cdot \frac{7(7 + 1)(2(7) + 1)}{6} = 700$$ For the next part, let's use property #1: $$\sum_{i=1}^{n}c=cn$$ In this case, c = 2 and n = 7. Let's plug in and simplify. $$\sum_{i=1}^{7}2= 2 \cdot 7 = 14$$ Now we can put the two together and find our sum. $$\sum_{i=1}^{7}(5i^2 + 2)= 700 + 14 = 714$$

When the Lower Limit of Summation is not 1

In the typical algebra tutorial, we are met with a definition for sequences in which the natural numbers are the domain. As we progress into Calculus and higher-level math, we might see a more general definition of a sequence that includes a domain as a subset of the integers. In this case, you might get index values of negative integers or zero. At this point, we are really not going to work with such definitions. But it is useful to see a few problems in which the lower limit of summation is not 1. This will require us to do some additional work beyond just the typical plug and chug with our given formulas. Let's look at a few examples.
Example #4: Evaluate each series. $$\sum_{i = 0}^{5}(i + 600)$$ Let's rewrite our series using the properties of summation. $$\sum_{i = 0}^{5}(i + 600) = \sum_{i = 0}^{5}i + \sum_{i = 0}^{5}600$$ First Part (leftmost): $$\sum_{i = 0}^{5}i = 0 + \sum_{i = 1}^{5}i = \frac{5(5 + 1)}{2} = 15$$ The first iteration (where i = 0) produces a value of 0. Since adding 0 doesn't change anything, we don't have any change when compared to our usual scenario when i = 1. You can see this more clearly if you work the problem the long way. $$\sum_{i = 0}^{5}i = {\color{red} 0} + 1 + 2 + 3 + 4 + 5 = 15$$ Second Part (rightmost): $$\sum_{i = 0}^{5}600 = 600 + \sum_{i = 1}^{5} 600 = 600 + 600(5) = 600 + 3000 = 3600$$ Since i = 0 instead of i = 1, we have an additional iteration when finding our sum. This adds an additional 600 to the result of the usual scenario when i = 1. $$\sum_{i = 0}^{5}(i + 600) = 15 + 3600 = 3615$$ Example #5: Evaluate each series. $$\sum_{i = -1}^{4}(i^2 + 3i + 10)$$ Let's rewrite our series using the properties of summation: $$\sum_{i = -1}^{4}(i^2 + 3i + 10) = \sum_{i = -1}^{4}i^2 + 3\sum_{i = -1}^{4} i + \sum_{i = -1}^{4} 10$$ Since i = -1, we will have two additional iterations (i = -1 and i = 0) to consider.
First Part (leftmost): $$\sum_{i = -1}^{4}i^2 = (-1)^2 + 0^2 + \sum_{i = 1}^{4} i^2 = 1 + \frac{4(4 + 1)(2(4) + 1)}{6}= 1 + \frac{4(5)(9)}{6} = 1 + 30 = 31$$ Second Part (middle): $$3\sum_{i = -1}^{4} i = 3\left(-1 + 0 + \sum_{i = 1}^{4} i\right) = 3\left(-1 + \frac{4(4 + 1)}{2}\right) = 3(-1 + 10) = 3(9) = 27$$ Last Part (rightmost): $$\sum_{i = -1}^{4} 10 = 10 + 10 + \sum_{i = 1}^{4} 10 = 20 + 10(4) = 20 + 40 = 60$$ Putting everything together gives us our sum. $$\sum_{i = -1}^{4}(i^2 + 3i + 10) = 31 + 27 + 60 = 118$$

Re-Indexing the Sum (Optional Reading)

For the previous two examples, the lower limit of summation was not 1. If we are given simple examples where the lower limit of summation is set equal to 0 or -1, it really isn't that much extra work to do the additional calculations beyond the standard formula (used with i = 1). But what if you were given a lower limit of summation of something like: 5, 13, or even 40? What would you do in that situation? Well, it turns out we have this concept of re-indexing the sum that is quite useful for this situation. Again, you probably won't see any examples until Calculus where this is needed, but we can think about it with a simple example.
Example #6: Evaluate each series. $$\sum_{i = 6}^{15} (i^2 + i + 5)$$ We want to adjust the starting index from 6 to 1. We can use a simple substitution to accomplish this task.
Since 6 - 5 = 1: $$\text{let} \, j = i - 5$$ $$\text{then} \, i = j + 5$$ Let's replace the i = 6 with j = 1. The upper limit of summation is now 10 (15 - 5 = 10). Since things are in terms of j now, we need to replace each i with (j + 5). $$\sum_{j = 1}^{10} ((j + 5)^2 + (j + 5) + 5)$$ Simplify: $$\sum_{j = 1}^{10} (j^2 + 10j + 25 + j + 5 + 5) = \sum_{j = 1}^{10} (j^2 + 11j + 35)$$ Let's rewrite the series using the summation properties: $$\sum_{j = 1}^{10} (j^2 + 11j + 35) = \sum_{j = 1}^{10} j^2 + 11\sum_{j = 1}^{10} j + \sum_{j = 1}^{10} 35$$ First Part (leftmost): $$\sum_{j = 1}^{10} j^2 = \frac{10(10 + 1)(2(10) + 1)}{6} = \frac{10(11)(21)}{6} = 385$$ Second Part (middle): $$11\sum_{j = 1}^{10} j = 11 \cdot \frac{10(10 + 1)}{2} = 11 \cdot \frac{10(11)}{2} = 605$$ Last Part (rightmost): $$\sum_{j = 1}^{10} 35 = 35(10) = 350$$ $$\sum_{j = 1}^{10} ((j + 5)^2 + (j + 5) + 5) = 385 + 605 + 350 = 1340$$ Let's write our sum in terms of the original problem: $$\sum_{i = 6}^{15} (i^2 + i + 5) = 1340$$

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