\documentclass[11pt]{article}
\usepackage{precalc}
\begin{document}
\assigntitle{24}{Series}
\section{Sigma notation}
\topic{series}
A \term{series} is the sum of the terms in some sequence. For
example, if we have the simple arithmetic sequence $1$, $2$, $3$, $4$, $5$,
$\dots$, the corresponding series is $1 + 2 + 3 + 4 + 5 + \dots$ and so
on. Of course, adding up this particular series is meaningless, since
it goes on forever. But it is often useful to talk about
\term{partial sums} of a series, where we add up some of the terms and
then stop. In general, we will use the notation $S_n$ to denote the
sum of the first $n$ terms of a series. For example, for the series
just mentioned, $S_4$ would be equal to $1 + 2 + 3 + 4 = 10$.
\begin{problem}
Consider the arithmetic sequence with $a = 5$ and $d = 2$. What is $S_6$?
\end{problem}
\topic{$\Sigma$ notation}
\term{Sigma notation} can be used to concisely represent a series.
Instead of explaining how it works, first I'll just give some examples:
\begin{align*}
\sum_{k=1}^4 k &= 1 + 2 + 3 + 4 = 10 \\
\sum_{k=1}^4 2k &= 2\cdot 1 + 2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4 = 2
+ 4 + 6 + 8 = 20 \\
\sum_{j=5}^7 j^2 &= 5^2 + 6^2 + 7^2 = 25 + 36 + 49 = 110 \\
\begin{split}
\sum_{n=-5}^{-2} \frac{2n+1}{n^2} &= \frac{2 \cdot (-5) + 1}{(-5)^2}
+ \frac{2 \cdot (-4) + 1}{(-4)^2} + \frac{2 \cdot (-3) + 1}{(-3)^2}
+ \frac{2 \cdot (-2) + 1}{(-2)^2} \\
&= -\frac{9}{25} - \frac{7}{16} -
\frac{5}{9} - \frac{3}{4} = -\frac{7571}{3600}
\end{split}
\end{align*}
Got that? The big thing that looks kind of like an \emph{E} is not an
E at all; it's a capital sigma, the 18th letter of the Greek
alphabet.\footnote{Sigma corresponds to our letter S, which stands for
\emph{sum}.} The variable at the bottom of the sigma is the
\emph{index of summation}, and the numbers at the bottom and top are
the \emph{limits of summation}. $\sum_{k=a}^b f(k)$ can be read as
``the sum, as $k$ goes from $a$ to $b$, of $f(k)$.'' The idea is that
you take each value of $k$ starting from $a$ and counting \emph{by
ones} up to (and including) $b$, and substitute it into the
right-hand formula; then you add up everything you got by doing the
substitutions.
You can create sigma notation in \LaTeX\ using the \verb|\sum|
command, followed by a subscript and a superscript for the parts below
and above the $\Sigma$. For example, \verb|\sum_{k=1}^{20} (k^2 + 3)|
produces \[ \sum_{k=1}^{20} (k^2 + 3). \]
\newcommand{\ds}{\displaystyle}
\begin{problem} \label{prob:try-sums}
Now you try a few.
\begin{subproblems}
\item $\ds \sum_{k=1}^5 (2k-1)$
\item $\ds \sum_{j=1}^6 j^2$
\item $\ds \sum_{r=2}^2 r\pi$
\item $\ds \sum_{g=5}^2 (g + 6)$
\item $\ds \sum_{j=1}^4 \left( \sum_{k=2}^3 jk \right)$
\end{subproblems}
\end{problem}
Series are extremely common, so it's important to be able to evaluate,
use, and manipulate them comfortably. Sigma notation allows us to be
precise about series, and to be able to see the ``big picture''
without getting lost in scads of ellipses. Writing \[ \sum_{k=1}^{10}
k \] is not only more concise than writing something like \[ 1 + 2 + 3
+ \dots + 10, \] but it can also be much easier to read and work with,
once you get used to it.\footnote{It's also much easier to
impress your friends with!}
\section{Some special series}
\label{sec:special-series}
\begin{problem} \label{prob:sum-of-ones}
What is \[ \sum_{k=1}^n 1 \quad \text{?} \]
\end{problem}
\begin{problem} \label{prob:triangular-nums}
\topic{arithmetic series}
This problem will guide you through finding a \term{closed form} (by
which, in this context, I mean an equivalent expression that does
not involve a sum) for the series \[ \sum_{k=1}^n k, \] that is, the
sum of the first $n$ whole numbers. As you may remember from a few
assignments ago, these partial sums ($1$, $1+2$,
$1+2+3$, $\dots$) are called the \term{triangular numbers}.
Let $S = \sum_{k=1}^n k = 1 + 2 + 3 + \dots + n$; we want to find a
nice expression for $S$ in terms of $n$. Consider adding $S$ to
itself, but pair $1$ with $n$, $2$ with $(n-1)$, and so on; each
pair of numbers will add up to $(n+1)$. Like this:
{\setlength\arraycolsep{2pt}
\[ \begin{array}{*{12}c}
& S & = & 1 & + & 2 & + & 3 & + & \cdots & + & n \\
+ & S & = & n & + & (n-1) & + & (n-2) & + & \cdots & + & 1 \\
\hline
& 2S & = & (n+1) & + & (n+1) & + & (n+1) & + & \cdots & + & (n+1)
\end{array} \] }
\begin{subproblems}
\item Use this last equation to solve for $S$. (\emph{Hint:} how many
copies of $(n+1)$ are there?
\item Use your solution for $S$ to find the $327$th triangular
number.
\end{subproblems}
\end{problem}
\begin{problem} \label{prob:geometric-sum} \topic{geometric series}
This problem will guide you
to finding a closed form for the sum \[ \sum_{k=0}^n r^k. \]
Let $S = r^0 + r^1 + r^2 + \dots + r^n$.
\begin{subproblems}
\item What is $r \cdot S$?
\item What is $S - r \cdot S$? Simplify your answer as much as
possible (hint: a lot of things should cancel each other).
\item Note that you can factor $S - r \cdot S$ as $S (1 - r)$.
From the previous part you should now have an equation that
looks like $S (1 - r) = \dots$ where the dots are your answer to
part (b). Now divide both sides by $(1-r)$ to obtain a closed
form for $S$.
\end{subproblems}
\end{problem}
\begin{problem}
What is \[ \sum_{i=0}^8 2^i? \]
\end{problem}
\begin{problem} \label{prob:geometric-polynomial}
Simplify: $2x + 4x^2 + 8x^3 + 16x^4 + 32x^6$.
\end{problem}
\section{Properties of sigma notation}
\label{sec:properties}
It's also important to investigate a few general properties of sigma
notation, and see how we can use these properties to manipulate series.
Suppose that $f(k)$ and $g(k)$ are any functions of $k$, and $c$ is
any constant (that is, an expression that does not involve $k$---in
particular, it doesn't have to be just a number; it could involve
\emph{other} variables so long as it does not involve $k$). Then the
following properties always hold:
\begin{align}
&\sum_{k=a}^b c = (b - a + 1) \cdot c && \text{(sum of constants)} \label{eq:sigma-constant-sum} \\
&\sum_k c \cdot f(k) = c \sum_k f(k) && \text{(distributivity)\footnotemark}\label{eq:sigma-constant} \\
&\sum_k \left[f(k) \pm g(k) \right] = \sum_k f(k) \pm \sum_k g(k) &&
\text{(associativity)}
\label{eq:sigma-sum}
\end{align}
\footnotetext{By the way, the
fact that there are no limits of summation in this equation and the
next isn't a typo; it just means that these equations hold
independently of any particular limits of summation.}
These might look scary, but they're actually quite simple. The first
property (\pref{eq:sigma-constant-sum}) expresses what happens when we
add up a constant expression: since it doesn't involve $k$, it doesn't
change, and we just add up $(b-a+1)$ copies of it (you should convince
yourself that there are $(b-a+1)$ numbers from $a$ to $b$ inclusive).
For example, \[ \sum_{k=2}^5 4e = 4e + 4e + 4e + 4e = 16e. \]
\begin{problem}
What is \[ \sum_{j=9}^{20} 6? \]
\end{problem}
The second property (\pref{eq:sigma-constant}) says that we can factor
any constant expression out of a sum: this is just the normal
distributive property of multiplication over addition. For example,
saying that $(2 + 4 + 6 + 8 + 10) = 2 \cdot (1 + 2 + 3 + 4 + 5)$ is
the same as saying that \[ \sum_{k=1}^5 2 k = 2\sum_{k=1}^5 k. \]
The last property (\pref{eq:sigma-sum}) says that we can break up a
sum of sums (or differences) into the sum (or difference) of separate
sums. Really, all this says is that we can change the order of
addition without affecting the result. For example, $(1 + 5) + (2 +
6) + (3 + 7) = (1 + 2 + 3) + (5 + 6 + 7)$; equivalently, we can say
that \[ \sum_{k=1}^3 (k + (k+4)) = \sum_{k=1}^3 k + \sum_{k=1}^3
(k+4). \]
However, you should note that the corresponding property does \emph{not} hold
for multiplication:
\begin{equation} \label{eq:sum-of-products}
\sum_k f(k) \cdot g(k) \neq \left( \sum_k f(k) \right) \cdot \left(
\sum_k g(k) \right).
\end{equation}
That is, a sum of products is \emph{not} the same thing as a product
of sums!
\begin{problem}
Why not? Give an example to illustrate~\pref{eq:sum-of-products}.
\end{problem}
These properties are fairly simple, but in conjunction with a few
special series we know
(Problems~\ref{prob:sum-of-ones}--\ref{prob:geometric-sum}), we can
already use them to tackle some more complex series. For example:
\begin{align*}
\sum_{k=1}^n (2k + 3) &= \sum_{k=1}^n 2k + \sum_{k=1}^n 3 &&
\text{property \eqref{eq:sigma-sum}} \\
&= 2 \sum_{k=1}^n k + \sum_{k=1}^n 3 && \text{property
\eqref{eq:sigma-constant}} \\
&= 2 \cdot \frac{n (n+1)}{2} + 3n &&
\text{\pref{prob:triangular-nums}, property \eqref{eq:sigma-constant-sum}} \\
&= n^2 + n + 3n = n^2 + 4n. && \text{simplify}
\end{align*}
\begin{problem} \label{prob:sum-arith-3-2}
Let $Q$ represent the arithmetic sequence with $a=3$ and $d=2$. Find
a closed-form expression for the sum of the first $n$ terms of $Q$.
(\emph{Hint:} remember, the $k$th term of $Q$ can be written as $a + (k-1)d = 1
+ 2k$. Write down the appropriate sum and then use sigma notation
properties to simplify, as in the example above.)
\end{problem}
\end{document}