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 April 3rd, 2007, 10:16 PM #1 Newbie   Joined: Apr 2007 Posts: 3 Thanks: 0 Calc 2, Comparison Tests, need help using the Calculator A question is asking me to approximate the first 100 terms of [Riemann sum (n=1) 1/(n^3)+1]. I'm not sure how to use my calculator in order to figure out the sum of the series. April 4th, 2007, 01:01 AM #2 Site Founder   Joined: Nov 2006 From: France Posts: 824 Thanks: 7 What do you call a Riemann sum ? Are you simply trying to evaluate sum(1+1/n^3) ? I've always heard of Riemann sums in a context of Riemann-integration ... April 4th, 2007, 02:10 AM #3 Newbie   Joined: Apr 2007 Posts: 3 Thanks: 0 Here is the question: Use the sum of the first 100 terms to approximate the sum of the series ∑(n=1, n=∞) 1/(n^3)+1. Estimate the error involves in this approximation. To start off I concluded that: the given series is convergent since 1/(n^3)+1 < 1/(n^3) by the comparison test. After continuing this problem, I'm up to the step where I have to solve ∑(n=1, n=100) 1/(n^3)+1, but I don't know how to do it. I could plug in each of the 100 terms and add them up, but I know there is a way to do it using the calculator. April 4th, 2007, 04:51 AM #4 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Do you mean 1/(n^3)+1 or 1/((n^3)+1) ? 1/(n^3)+1 diverges by the nth term test. lim n → ∞ 1/(n³)+1 = lim n → ∞ ((n³) + 1)/(n³) = 1 =diverges. 1/((n³)+1), however, converges by the comparison test with 1/n³ as you say. Unfortunately, I don't know how to get a calculator to approximate this sort of problem, but I do know how to write programs for my computer to approximate almost anything: Code: import java.util.*; public class approximate { public static void main (String[] args) { float answer = 0; for(int n=1; n<=100; n++) { answer += 1/(Math.pow(n,3)+1); } System.out.println(answer); } } Approximate answer after 100 terms = 0.6864537 April 10th, 2007, 01:33 PM #5 Newbie   Joined: Apr 2007 Posts: 3 Thanks: 0 Thank you  April 10th, 2007, 04:01 PM   #6
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