\problemdata      
        {?} % latex-friendly label for the prob.
        {Summations} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, PS7; Spring 98, PS5} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
Let $S_n = \sum_{i=1}^n i^{1 /3}$.  Use the integral method to determine
\[
\lim_{n \rightarrow \infty} \frac{S_n}{n^{4 / 3}}.
\]

\solution{
We use the integral method for bounding $S_n$. 
\[\int_0^n x^{\frac{1}{3}}dx \leq S_n \leq \int_1^{n+1}x^{\frac{1}{3}}dx.\]
That is,
\[\frac{(3/4)n^{4/3}}{n^{4/3}}  \leq \frac{S_n}{n^{4/3}} \leq   \frac{(3/4)((n+1)^{4/3} - 1)}{n^{4/3}}\]
Taking limits as $n \rightarrow \infty$ we find that
\[3/4 \leq {\displaystyle \lim_{n \rightarrow \infty}\frac{S_n}{n^{4/3} }} \leq 3/4\]

Therefore, ${\displaystyle \lim_{n\rightarrow\infty}\frac{S_n}{n^{4/3}}} = 3/4.$
}
\end{problem}
