Chapter 9.4, Problem 39ES

Exercise 36 will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises 37-40.

It was stated in Exercise 35 that ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{k^4}=\frac{{\pi }^4}{90}}$

(a) Let $s_n$ be the nth partial sum of the series above. Show that $s_n+\frac{1}{3{\left(n+1\right)}^3}<\frac{{\pi }^4}{90}<s_n+\frac{1}{3n^3}$

(b) We can use a partial sum of the series to approximate ${\pi }^4/90$ to three decimal-place accuracy by capturing the sum of the series in an interval of length 0.001 (or less). Find the smallest value of n such that the interval containing ${\pi }^4/90$ in part (a) has a length of 0.001 or less.

(c) Approximate ${\pi }^4/90$ to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of ${\pi }^4/90.$