\input epsf
\nopagenumbers
\magnification=\magstep1

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

\smallskip

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

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

\medskip

\noindent a) The series has all positive terms. Therefore if the
``infinite tail'' $\sum\limits_{j=16}^{\infty}{{30^j}\over{(j!)^2}}$
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 $a_j={{30^j}\over{(j!)^2}}$, 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$. (You'll need a calculator!)

\medskip

\noindent c) Assume (this is true!) that $a_{16}<.00099$. Use this
fact and what was done in c) 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. (Show all
steps. You'll need a calculator!)

\vfil\eject\end