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\begin{document}


\title{Math 21, Fall, 2000}


\address{Lehigh University}


\email{david.johnson@lehigh.edu}

\maketitle

\subsection{Sigma notation}

We will be spending quite a bit of time dealing with integration as a summation
process, and it will get awkward writing down all these sums. We need a better
notation. Usually the sums we will be ``casting up'' are \textit{indexed}, that
is, there is a whole number for each term in the sum, beginning with either
0 or 1 (sometimes higher), and ending at some larger value, or at infinity.
So, instead of writing \[
a_{1}+a_{2}+a_{3}+\dots +a_{n},\]
we will write \[
\sum _{i=1}^{n}a_{i},\]
meaning the same thing. We call the second line \textit{sigma notation}. Sometimes
we use \( k \) or \( j \) or some other letter instead of \( i \), but it
doesn't matter, since it's just a ``dummy index'' of summation, like the name
of a variable in the expression for a function --- it's not the name of the
variable, but the formula, that matters. The \( i \) or whatever often is involved
in the expression for \( a_{i} \), like a function. For example; let's look
at: \[
\sum _{j=1}^{4}\frac{1}{j}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}.\]


The sum \[
\sum _{k=1}^{4}\frac{1}{k}\]
 is \textbf{exactly} the same thing, \( 1/1+1/2+1/3+1/4=25/12 \). Other examples:
 

\begin{xca}
Compute \[
\sum _{i=1}^{3}2^{i}=\]
 
\end{xca}
%

\begin{xca}
Find \[
\sum _{n=1}^{10}1=\]
 
\end{xca}
Think about this for a bit.

\begin{xca}
Find \[
\sum _{k=2}^{6}(-1)^{k}=\]
 
\end{xca}
%

\begin{example}
Find \[
\sum _{i=2}^{100}(\frac{1}{i}-\frac{1}{i-1}).\]
 
\end{example}

\subsubsection*{Solution}


\subsection*{Summation formulas}

The way I heard the story, little Carl Gauss http://www-groups.dcs.st-and.ac.uk/\textasciitilde{}history/Mathematicians/Gauss.html
was a bit of a discipline problem in school. As punishment one day, his teacher
sent him to add the first thousand integers together. Little Carl came back
rather sooner than the teacher had expected, with an answer. The teacher thought
it was impossible to add that many numbers together that quickly. But, of course,
little Carl was no ordinary discipline problem. He had figured out the following
general formula: \[
\sum _{i=1}^{n}i=\frac{(n)(n+1)}{2},\]
or, without the sigma: \[
1+2+3+4+5+\dots +n=\frac{(n)(n+1)}{2},\]
and, since Gauss grew up to be one of the greatest mathematicians of all time,
we can suppose that he showed his teacher the right sum, 500,500. 

How did he figure that out? I suppose he used some sort of \textit{mathematical
induction}. He reasoned this way: he could easily show that the formula was
true for \( n=1 \). He could also show that, \textbf{\textit{if}} the formula
is true for some integer \( n \), \textbf{\textit{then}} it is true for the
next integer, which is \( n+1 \). Thus; it's true for 1. But, if it's true
for 1 it's true for 2. If it's true for 2 it's true for 3, \textit{etc}. Where
could it be false? It can't be false for any \( n \). That is the essence of
an induction proof; you need a place to start, where the statement is true,
and you need to be able to show the next step, \textit{if} you \textit{\emph{assume}}
that the previous level is true. 

Let's try it here: \[
\sum _{i=1}^{1}i=1,\]
which is the same as \[
1=\frac{(1)(1+1)}{2},\]
so the formula is true for \( n=1 \). Now, assume that, for a given \( n \),
that the formula is true: \[
\sum _{i=1}^{n}i=\frac{(n)(n+1)}{2}.\]
We have to show that it's true for the \textit{next} value of \( n \): \[
\sum _{i=1}^{n+1}i=\frac{(n+1)((n+1)+1)}{2}.\]


The way you do that is to look at the sum: \begin{eqnarray*}
\sum _{i=1}^{n+1}i & = & \left( \sum _{i=1}^{n}i\right) +(n+1)\\
 & = & \frac{n(n+1)}{2}+(n+1)\\
 & = & \frac{n(n+1)}{2}+\frac{2(n+1)}{2}\\
 & = & \frac{n(n+1)+2(n+1)}{2}\\
 & = & \frac{(n+2)(n+1)}{2}\\
 & = & \frac{(n+1)((n+1)+1)}{2},
\end{eqnarray*}
which is the same formula, but for \( n+1 \). As little Gauss did, put it together
and it says that that formula is true for all \( n \). 

There are two other formulas like this that come in handy: \[
\sum ^{n}_{i=1}i^{2}=\frac{n(n+1)(2n+1)}{6},\]


and \[
\sum ^{n}_{i=1}i^{3}=\left[ \frac{n(n+1)}{2}\right] ^{2},\]
which can be proved the same way.  There is a formula like this for the sum
of the \( k^{th} \) powers of the first \( n \) integers, for any \( k \),
but these are the only ones commonly used. The one pattern that continues for
all \( k \) is that the formula always involves the \( (k+1)^{st} \) power
of \( n \). The curious coincidence between the first powers and the third
powers is not, so far as I know, repeated. 

Of course, with modern computers we can all add more rapidly than Gauss' teacher.
But these general formulas still play a significant role, as we'll see. 

\begin{example}
Find \[
\sum _{i=1}^{n}\frac{3}{n}\left( \left( \frac{i}{n}\right) ^{2}+1\right) .\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Find \[
\sum _{j=1}^{n}\left( j^{2}+3j+1\right) =\]

\end{xca}
\begin{example}
Evaluate \[
\sum _{i=1}^{m}\left( \sum _{j=1}^{n}\left( i+j\right) \right) .\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Find \[
\lim _{n\rightarrow \infty }\left( \sum _{i=1}^{n}\frac{1}{n}\left( \frac{i}{n}\right) ^{2}\right) =\]
\end{xca}

\end{document}
