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\begin{document}

\tutn{12}{10/22/99}{October 24, 1999}

\begin{problems}

\problem Is a Harvard degree really worth more than an MIT degree?!  Let
us say that a person with a Harvard degree starts with \$40,000 and gets a
\$20,000 raise every year after graduation, whereas a person with an MIT
degree starts with \$30,000, but gets a 20\% raise every year.  Assume
inflation is a fixed $8\%$ every year.  That is, \$1.08 a year from now is
worth \$1.00 today.  (You'll need a calculator to get final answers; if
one is not available, it's ok to express the answer as a closed form
numerical expression.)

\bparts


\ppart How much is a Harvard degree worth today if the holder will work
for $n$ years following graduation? \hint In Lecture 8 a closed formula
for $\sum_{i=0}^{n} ir^i$ was derived.

\solution{

One dollar after year $i$ is worth $r^i$ in today's currency, where
\[
r= \frac{1}{1.08} = 0.925\, 925\, 925 \dots.
\]
So 
\begin{eqnarray*}
\mbox{Hvd}_n & = & \sum_{i=0}^{n} (40000+ 20000i)r^i\\
        & = & 40000\sum_{i=0}^{n} r^i + 20000\sum_{i=0}^{n} i r^i.
\end{eqnarray*}

But we know from Lecture Notes 8 that
\[
\sum_{i=0}^{n} ir^i = \frac{{r - (n+1) r^{n+1} + n r^{n+2}}}{(1-r)^2},
\]
so 
\begin{equation}
\mbox{Hvd}_n  =  40000 \frac{(1- r^{n+1})}{1 - r}  + 
                   20000 \frac{(r - (n+1) r^{n+1} + n r^{n+2})}{(1-r)^2},
\end{equation}
and for $n=20$,
\begin{equation}
\mbox{Hvd}_{20}  = \$2,010,885.
\end{equation}
}

\ppart How much is an MIT degree worth in this case?

\solution{

\begin{eqnarray*}
\mbox{MIT}_n & = & 30000\sum_{i=0}^{n} {1.2}^i r^i\\
        & = & 30000\sum_{i=0}^{n} (1.2r)^i\\
        & = &  \frac{{30000(1 - (1.2r)^{n+1})}}{1 - 1.2r}
\end{eqnarray*}
and for $n=20$,
\begin{eqnarray*}
\mbox{MIT}_{20} & = &  \frac{{30000(1 - (1.2r)^{21})}}{1 - 1.2r} = 2,197,579.
\end{eqnarray*}

}

\ppart If you plan to retire after twenty years, which degree would
be worth more?

\solution{

The MIT degree is more valuable---as if you couldn't guess \texttt{:-)}
---but only by about 5\%.  But if you were persistent enough to work 30
years, you would find the MIT degree was worth more than twice as much.

}

\eparts


\problem

Regrouping or reordering the terms of an infinite series is ok for
\emph{absolutely convergent} series, namely, series for which the sum of
the absolute values of their terms converges.

In other cases, rearranging terms is not ok.
An extreme example is
\[
S \eqdef \sum_{i=0}^{\infty} (-1)^i = 1 + (-1) + 1 + (-1) + \cdots
\]
because by regrouping the terms we can deduce:
\begin{eqnarray*}
S & = & [ 1 + (-1) ] + [ 1 + (-1) ] + \cdots = 0 + 0 + \cdots = 0,\\
S & = & 1 + [(-1) + 1] + [(-1) + 1] + \cdots = 1 + 0 + 0 + \cdots = 1.
\end{eqnarray*}

To see the problem here, let $S_n \eqdef \sum_{i=0}^{n}(-1)^i$.  Then $S_n
= 1$ for odd $n$ and $S_n = 0$ for even $n$.  Hence $S_n$ does not have a
well-defined limit as $n$ tends to infinity, implying that $S$ is not a
convergent series.

Even for convergent, but not absolutely convergent, series, rearranging
terms can cause big changes.  For example, the series
\[
1 - 1/2 + 1/3 - 1/4 \dots
\]
converges, but is not absolutely convergent.

Prove that by reordering terms in $1 - 1/2 + 1/3 - 1/4 \dots$, you can
obtain three series which converge to 3, converge to 4, and don't converge.

\solution{

Note that the sum of the negative terms goes to negative infinity, because
the sum
\[
-1/2 - 1/4 - 1/6 -  \dots - 1/2n = -H_n/2 \sim (-\ln n)/ 2
\]
and $\ln n$ goes to infinity with $n$.  Similarly, the sum of the positive
terms goes to positive infinity.

To reorder the terms to converge to 3, begin by taking the positive terms
in decreasing order until their sum first exceeds 3.  Next, take the
negative terms in order of decreasing magnitude until the sum of all the
terms taken---positive and negative---first goes below 3.  Now take the
remaining positive terms in decreasing order until the sum again exceeds 3,
then the negative terms till the sum goes below 3, and so on.  It is always
possible to move the sum above or below 3 by taking enough of the remaining
positive or negative terms, because the remaining positive terms sum to
positive infinity and the remaining negative terms sum to negative
infinity.  Also, the running sum of the positive and negative terms taken
in this way goes above and below 3 at most by the magnitude of the last
term taken.  Since the magnitude of the terms goes to zero, this implies
that the sum goes above and below 3 by an amount that decreases to zero.
But this is exactly what it means to say the sum converges to 3.

Replacing ``3'' by ``4'' above yields an ordering in which the sum
converges to 4.

Finally, alternately taking enough positive terms to increase the sum by 1
and then taking one negative term, yields an ordering in which the sum
diverges to positive infinity.

This argument applies to \emph{any} convergent series that is not
absolutely convergent to show that it can be reordered to converge to
\emph{any} desired number, or to diverge.  The moral: when reordering terms
in a series, be sure to check for absolute convergence.

}

\problem
There is a number $a$ such that $\sum_{i=1}^\infty i^p$ converges iff $p < a$.

\bparts

\ppart What is the value of $a$?

\solution{$-1$}

\ppart Prove it.

\solution{

For $p=-1$, the sum is the harmonic series which we know does not converge.
Since the term $i^p$ is increasing in $p$ for $i>1$, the sum will be
larger, and hence also diverge for $p>-1$.

For $p<0$, the value of $i^p$ decreases as $i$ increases, so by the
integral method, the sum is bounded above by one plus the integral from 1
to $\infty$ of $x^p$.  The indefinite integral is $x^{p+1}/(p+1)$ and so
for $p < 1$, the infinite integral is $1/(p+1)$.  Hence the sum is bounded
above, and since it is increasing, it has a finite limit, i.e., it
converges.

}

\eparts


\end{problems}

\end{document}




