\problemdata     
{S01-PS5-3}           
{Integral method}
{}              
{F98-PS5, S01-PS5} 		
{S01 Karger}

\begin{problem}


For the following summation use the integral method to find
upper and lower bounds that differ by at most 0.1.  Note that you may
need to add the first few terms explicitly and then use integrals to
bound the sum of the remaining terms.

\[
\sum_{i=1}^{\infty} \frac{1}{i^2}
\]

\solution {

We can upper bound the summation as follows:

\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{1}{i^2}
    & \leq &    \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \int_3^{\infty} \frac{1}{x^2} \ dx \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \left( - \frac{1}{x} \right)_3^{\infty} \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \frac{1}{3} \\
    & = &       1.694\ldots
\end{eqnarray*}

We can lower bound the summation similarly:

\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{1}{i^2}
    & \geq &    \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \int_3^{\infty} \frac{1}{(x+1)^2} \ dx \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \left( - \frac{1}{x+1} \right)_3^{\infty} \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \frac{1}{4} \\
    & = &       1.611\ldots
\end{eqnarray*}
}

(The actual value of the summation turns out to be $\pi^2 / 6 =
1.644\ldots$.)  

\end{problem}