Homework:HW SECTION 10.4 Question 5, 10.4.41 Part 4 of 4 HW Score: 85.71%, 6 of 7 points Points: 0 of 1

Skip to Main content Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question content area top Part 1 Consider the following convergent series. Complete parts a through d below. Summation from k equals 1 to infinity StartFraction 9 Over 8 k Superscript 6 EndFraction Question content area bottom Part 1 a. Use an integral to find an upper bound for the remainder in terms of n. The upper bound for the remainder is    StartFraction 9 Over 40 n Superscript 5 EndFraction .    Part 2 b. Find how many terms are needed to ensure that the remainder is less than 10 Superscript negative 3. The minimum number of terms needed is    3. ​(Round up to the nearest whole number. Use the answer from part a to answer this​ part.) Part 3 c. Use an integral to find lower and upper bounds ​(Upper L Subscript n and Upper U Subscript n ​respectively) on the exact value of the series. Upper L Subscript nequalsUpper S Subscript n Baseline plus    StartFraction 9 Over 40 left parenthesis n plus 1 right parenthesis Superscript 5 EndFraction and Upper U Subscript nequalsUpper S Subscript n Baseline plus    StartFraction 9 Over 40 n Superscript 5 EndFraction ​(Type expressions using n as the​ variable.) Part 4 d. Find an interval in which the value of the series must lie if you approximate it using ten terms of the series. Using ten terms of the​ series, the value lies in the interval    enter your response here. ​(Do not round until the final answer. Then round to nine decimal places as needed. Use the answer from part c to answer this​ part.) The upper bound for the remainder is.The minimum number of terms needed is(Round up to the nearest whole number. Use the answer from part a to answer this part.)Upper L Subscript nUpper S Subscript n Baseline plusUpper U Subscript nUpper S Subscript n Baseline plusand Upper U Subscript nUpper S Subscript n Baseline plus(Type expressions using n as the variable.)Using ten terms of the series, the value lies in the interval(Do not round until the final answer. Then round to nine decimal places as needed. Use the answer from part c to answer this part.) Get more help