Sum's Upper Lower Bounds
Let \(S\ =\ \sum\limits_{n=1}^{57} \frac{1}{\sqrt[3]{n+7}}\). The Integral Method provides the lower and upper bounds on \(S\). By how much does the upper bound differ from the lower bound?
\(\frac{1}{\sqrt[3]{n+7}}\) is strictly decreasing over \(n = 1, 2, 3,..., 57 \). The lower bound is \(I+f(57) = I + \frac{1}{\sqrt[3]{57+7}} = I + \frac{1}{4}\). The upper bound is \(I+f(1) = I + \frac{1}{\sqrt[3]{1+7}} = I + \frac{1}{2}\). So the bounds differ by \((I + \frac{1}{2}) - (I + \frac{1}{4}) = \frac{1}{4}\). The value of \(I\) is not important to the question, but it equals the difference between \((3/2)((n+7)^{2/3})\) evaluated at 57 and at 1, namely, \((3/2)[64^{2/3} - 8^{2/3}] = (3/2)[4^2 - 2^2] = 18\).