Sequences and Series
Worked Examples
Series
If you try to add up all the terms of a sequence, you get an object called a series. In order to discuss series, it's useful to use sigma notation, so we will begin with a review of that.Part 1: Sigma Notation
When adding many terms, it's often useful to use some shorthand notation. Let $a_{n}$ be a sequence of real numbers. We set $$\sum_{i=1}^{k} a_{i} = a_{1} + a_{2} + \cdots + a_{k}$$ Here we add up the first terms $a_{1}, a_{2}, \ldots a_{k}$ of the sequence. We can also start the sum at a different integer. $$\sum_{i=j}^{k} a_{i} = a_{j} + a_{j+1} + \cdots a_{k}$$Example. Let $a_{n} = \frac{1}{2^{n}}$. Express the sum of the first 100 terms of the corresponding series, using sigma notation.
Solution: $$ \sum_{i=1}^{100} \frac{1}{2^i} $$Part 2: Basics
A series is an infinite sum $a_{1} + a_{2} + a_{3} + \cdots$. We typically write such an object using sigma notation $$\sum_{i=1}^{\infty} a_{i}.$$ If you do not use sigma notation to express a series, it's very important to include the ``$+ \cdots$" to make sure the person reading understands that the sum goes on forever.If you have an explicit expression for the term $a_{i}$ you usually replacen $a_{i}$ with this expression when using sigma notation. For example if $a_{i} = \frac{1}{2^{i}}$, then the corresponding series is usually written $ \sum_{i=1}^{\infty} \frac{1}{2^{i}}$.
Example. Consider the series $\frac{2}{3} + \frac{4}{9} + \frac{6}{27} +\frac{8}{81} + \cdots$. Find an expression for the ith term $a_{i}$, and use it to write the series in sigma notation.
Solution: If you look at this series carefully you see that each term is given by a fraction: $a_{i} = \frac{b_{i}}{c_{i}}$. Let's look at $b_{i}$ and $c_{i}$ separately. The $b_{i}$'s give the list $2, 4, 6, 8, \ldots$, or the even natural numbers. So $b_{i} = 2i$. The $c_{i}$ give the list $3, 9, 27, 81, \ldots$ which is the geometric sequence $c_{i} = 3^{i}$.Putting this together gives $a_{i} =\frac{2i}{3^{i}}$.
Mini-Lecture.
Below is a mini lecture about series.
Part 3: Partial Sums
Given a series $a_{1} + a_{2} + a_{3} + \cdots = \sum_{i=1}^{\infty}$ we can form the ``$n$th partial sum," usually denoted $S_{n}$. As the name might suggest, the $n$th partial sum is obtained by taking the first $n$ terms and adding them up. More concretely $ S_{n} = \sum_{i=1}^{n} a_{i} = a_{1} + a_{2} + \cdots a_{n} $.Example. Consider the series $\sum_{i=1}^{\infty} \frac{1}{2^{i}}$. Find the first 5 partial sums. (I.e. find the partial sums $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$.)
Solution: For $S_{1}$ we get: $S_{1} = a_{1} = \frac{1}{2}$.For $S_{2}$ we get: $S_{2} = a_{1} + a_{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$.
For $S_{3}$ we get: $S_{3} = a_{1} + a_{2} + a_{3} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$.
For $S_{4}$ we get: $S_{4} = a_{1} + a_{2} + a_{3} + a_{4} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{15}{16}$.
For $S_{5}$ we get: $S_{5} = a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} = \frac{31}{32}$.
Part 4: Geometric Series
A series of the for $ \sum_{i}^{\infty} ar^{i} = ar + ar^{2} + ar^{3} + \cdots$ is called a geometric series. (This is a series where the terms $a_{i}$ that we are summing, form a geometric sequence.)Let's consider the $n$th partial sum of a geometric sequence; $S_{n} = ar + ar^{2} + \cdots + ar^{n}$, as well as $rS_{n} = r(ar + ar^{2} + \cdots + ar^{n}) = ar^{2} + ar^{3} + \cdots + ar^{n+1}$.
The difference between these two expressions is $S_{n} - rS_{n} = ar - ar^{n+1}$. We can rearrange this and solve for $S_{n}$: \begin{align*} (1-r) S_{n} &= ar( 1-r^{n}) \\ S_{n} &= ar \frac{1-r^{n}}{1-r} \end{align*} So for a geometric series we get the formula: $$S_{n} = ar \frac{1-r^{n}}{1-r}$$
Example. Consider the series $\sum_{i=1}^{\infty} \frac{1}{2^{i}}$. Find the 100th partial sum, using the formula. Find a formula for the $n$th partial sum. What happens to this expression as $n$ gets larger and larger?
Solution: For the 100th partial sum, using the formula above we get $$ S_{100} = \frac{1}{2} \frac{1-(\frac{1}{2})^{100}} {1-\frac{1}{2}} = 2\frac{1}{2} (1-\frac{1}{2^{100}}) = 1-\frac{1}{2^{100}}.$$ For the $n$th partial sum, we get $$ S_{n} = \frac{1 - (\frac{1}{2})^{n}} {1-\frac{1}{2}} = 2\frac{1}{2} (1-\frac{1}{2^{n}}) = 1-\frac{1}{2^{n}} = 1-\frac{1}{2^{n}} .$$ As $n$ gets larger and larger, we see that the partial sum is getting closer and closer to $1.$As the previous example suggest, we can use the following formula to evaluate a geometric series, when the ratio $r$ satisfies $|r| <1$. $$\sum_{i=1}^{\infty} ar^{i} = \frac{ar}{1-r} \text{ for } |r|<1$$