\input epsf
\nopagenumbers
\magnification=\magstep1

\noindent {\bf Problem statement} A computer reports the following
information:

\smallskip

\centerline{$\sum\limits_{j=1}^{10}{{30^j}\over{(j!)^2}}\approx
6963.86479\,;\quad \sum_{j=1}^{15}{{30^j}\over{(j!)^2}}\approx
6977.78140\,;\quad \sum_{j=1}^{20}{{30^j}\over{(j!)^2}}\approx
6977.78249\,.$}

\noindent This suggests that
$\sum\limits_{j=1}^{\infty}{{30^j}\over{(j!)^2}}$ converges and that
its value (to at least 2 decimal places) is 6977.78. Explain the
details in the following outline of a verification of this statement.
In what follows, $a_j={{30^j}\over{(j!)^2}}$.

\medskip

\noindent a) The series has all positive terms. Therefore if the
infinite tail $\sum\limits_{j=16}^{\infty}a_j$ converges and has sum
less than .002, the omitted terms after the first 15 of the whole
series won't matter to 2 decimal places.

\medskip

\noindent b) If $j$ is a positive integer, simplify the algebraic
expression ${{a_{j+1}}\over {a_j}}$. Use this to show that if $j\ge
16$, then ${{a_{j+1}}\over {a_j}}<0.11$. (Show all steps. You'll need
a calculator!)

\medskip

\noindent c) It is true that $a_{16}\approx .000983$. Use this fact
and what was done in b) to compare $\sum\limits_{j=16}^{\infty}a_j$ to
a geometric series, each of whose terms is individually larger than
this series. Find the sum of the geometric series, which should be
less than .002, so that the omitted infinite tail of the original
series is small enough. (Show all steps. You'll need a calculator!)

\vfil\eject\end