You may assume basic facts about geometric sums and series for Essay 2. Let $r$ be any real number and let $n$ be a non-negative integer.  The sum
\begin{equation}\label{EqGeoSum}
1 + r + r^2 + \cdots + r^n
\end{equation}
is a {\em geometric sum} and the infinite series
\begin{equation}\label{EqGeoSeries}
1 + r + r^2 + \cdots + r^n + \cdots
\end{equation}
is a {\em geometric series}.

Suppose further that $r \neq 1$. Then the geometric sum (\ref{EqGeoSum}) can be computed by the formula
$$
1 + r + r^2 + \cdots + r^n  = \frac{1 - r^{n+1}}{1 - r}.
$$
This fact, which you may assume, is easily proved proved by mathematical induction.

Now suppose that  $|r| < 1$. Then $\lim_{n \longrightarrow \infty}r^n = 0$ which means the geometric series (\ref{EqGeoSeries}) converges to $\displaystyle{\frac{1}{1 - r}}$ by the preceding equation. We write
\begin{equation}\label{EqGeomSeriesSum}
1 + r + r^2 + \cdots + r^n + \cdots = \frac{1}{1 - r}
\end{equation}
to indicate that the series converges and to designate the limit of the sequence of partial sums.

Your essay will involve the geometric series
\begin{equation}\label{EqOneHalf}
1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} + \cdots .
\end{equation}
Since $\displaystyle{|\frac{1}{2}| < 1}$, it follows by (\ref{EqGeomSeriesSum}) that (\ref{EqOneHalf}) converges and
$\displaystyle{
1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} + \cdots = 2.
}$
The DeLux blocks are cubes with side lengths
$\displaystyle{
1, \quad \frac{1}{2}, \quad \frac{1}{3}, \quad \frac{1}{5}, \quad \ldots
}$
Your essay involves analyzing the sum of their side lengths
$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}  + \frac{1}{9} + \cdots +  \frac{1}{16}+ \cdots .
$$
The preceding series is called the {\em harmonic series}. Think of the terms of the geometric series (\ref{EqOneHalf}) as markers for grouping terms of the harmonic series as follows:
\begin{equation}\label{EqHarmonic}
{\BF 1} + \frac{\BF 1}{\BF 2} + (\frac{1}{3} + \frac{\BF 1}{\BF 4}) + (\frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{\BF 1}{\BF 8}) + (\frac{1}{9} + \cdots +  \frac{\BF 1}{\BF 16}) + \cdots .
\end{equation}
We will find an overestimate and an underestimate for the sum of the terms in each of the parenthesized groups. You will see a pattern emegering in our calculations:
$$
1 = \frac{1}{2} + \frac{1}{2} > \frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2},
$$
$$
1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} >
\frac{1}{5} + \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}  = \frac{1}{2},
$$
$$
1 = \frac{1}{8} + \cdots + \frac{1}{8}  = 8(\frac{1}{8}) >  \frac{1}{9} + \cdots +  \frac{1}{16} > \frac{1}{16} + \cdots +  \frac{1}{16} = 8(\frac{1}{16}) =  \frac{1}{2},
$$
$$
\vdots
$$
Using (\ref{EqHarmonic}) and our underestimates, we see that
\begin{eqnarray*}
\lefteqn{1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8} + \frac{1}{8} + \frac{1}{9} + \cdots +  \frac{1}{16}+ \cdots} \\
& = & 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4})+ (\frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}) + (\frac{1}{9} + \cdots +  \frac{1}{16}) + \cdots \\
& > & 1 + \frac{1}{2} + \frac{1}{2} +\frac{1}{2} +\frac{1}{2} + \cdots  .
\end{eqnarray*}
Thus the partial sums of the harmonic series grow without bound which is expressed by
$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+ \cdots = \infty.
$$
 

Below is a formal proof of the the fact that the sums of terms in parenthesized groupings lie between $\displaystyle{\frac{1}{2}}$ and $1$. You should {\em not} include the proof in your essay; the mathematics of your essay is to be treated informally. Observe that the terms of a parenthesized group are given by $\displaystyle{\frac{1}{2^n+1}, \ldots, \frac{1}{2^{n+1}}}$ for some $n \geq 1$.
\begin{Lemma}\label{MemGroupSums}
Let $n$ be a positive integer. Then
$\displaystyle{
\frac{1}{2} < \frac{1}{2^n + 1} + \cdots + \frac{1}{2^{n+1}} < 1.
}$
\end{Lemma}

\pf
Since $2^{n+1} = 2^n + 2^n$ the sum in the statement of the lemma has $2^n$ terms. Each term has the form $\displaystyle{\frac{1}{2^n + \ell}}$ for some $1 \leq \ell \leq 2^n$ and thus  satisfies
$$
\frac{1}{2^{n+1}} \leq \frac{1}{2^n + \ell} \leq \frac{1}{2^n}.
$$
At least one of the terms is larger than $\displaystyle{\frac{1}{2}}$ and one at least one is smaller then $1$. Therefore
$$
\frac{1}{2} = 2^n(\frac{1}{2^{n+1}}) <  \frac{1}{2^n + 1} + \cdots + \frac{1}{2^{n+1}} < 2^n(\frac{1}{2^n}) = 1.
$$
\qed
\vfill

DER 10/08/02
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