# 3.1 Sums & Products

## 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?

Exercise 1

$$\frac{1}{\sqrt[3]{n+7}}$$ 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, $$(3/2)[64^{2/3} - 8^{2/3}] = (3/2)[4^2 - 2^2] = 18$$.