\problemdata      
        {summations-3} % latex-friendly label for the prob.
        {Summations} % The topic(s) of the problem content.
        {S99 Q1} % Source (if known)
        {S02 CP6F; F01, PS7; S99 Q1} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem}
There is a number $a$ such that $\sum_{i=1}^\infty i^p$ converges iff $p <
a$.  What is the value of $a$?  Prove it.

\solution{$a= -1$.

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 $p>-1$.

By the integral method, the sum is $\Theta$ of the integral from
1 to $\infty$ of $x^p$.  For $p<-1$, the indefinite integral is $x^{p+1}/(p+1) =
\Theta(1/x^\epsilon)$ for $\epsilon = -1 - p > 0$, so the integral
evaluates to a constant.  Hence the sum is bounded above, and since it is
increasing, it has a finite limit, \ie it converges.}
\end{problem}
