Estimate \(\pi^2/6=\sum_{n=1}^{\infty}1/n^2\) by the sum of the first \(N\) terms. What is the smallest \(N\) such that the error is at most \(10^{-6}\)? Use the integral test.
Solution
The integral test shows that \(\begin{array} { r } { 1 / ( N + 1 ) < \sum _ { n = N + 1 } ^ { \infty } 1 / n ^ { 2 } < 1 / N . } \end{array}\) , so \(N = 1 0 ^ { 6 }\)